https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=StellarG&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-22T20:24:44Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_11&diff=141028 2017 AMC 8 Problems/Problem 11 2020-12-30T15:17:10Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 11==<br /> A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the number of tiles lying on both diagonals is &lt;math&gt;37&lt;/math&gt;, counting one tile twice, there are &lt;math&gt;37=2x-1\implies x=19&lt;/math&gt; tiles on each side. Hence, our answer is &lt;math&gt;19^2=361=\boxed{\textbf{(C)}\ 361}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> Associated video: https://youtu.be/QCWOZwYVJMg<br /> <br /> ==See Also:==<br /> {{AMC8 box|year=2017|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_12&diff=141027 2017 AMC 8 Problems/Problem 12 2020-12-30T15:16:38Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 12==<br /> The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder. The &lt;math&gt;\operatorname{LCM}(4,5,6)&lt;/math&gt; is &lt;math&gt;60&lt;/math&gt;. Since &lt;math&gt;60+1=61&lt;/math&gt;, and that is in the range of &lt;math&gt;\boxed{\textbf{(D)}\ \text{60 and 79}}.&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/SwgXyYFIuxM<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=11|num-a=13}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=139084 2019 AMC 8 Problems/Problem 25 2020-12-06T01:13:38Z <p>StellarG: /* Videos explaining solution */</p> <hr /> <div>==Problem 25==<br /> Alice has &lt;math&gt;24&lt;/math&gt; apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 1==<br /> We use [[stars and bars]]. Let Alice get &lt;math&gt;k&lt;/math&gt; apples, let Becky get &lt;math&gt;r&lt;/math&gt; apples, let Chris get &lt;math&gt;y&lt;/math&gt; apples.<br /> &lt;cmath&gt;\implies k + r + y = 24&lt;/cmath&gt;We can manipulate this into an equation which can be solved using stars and bars.<br /> <br /> All of them get at least &lt;math&gt;2&lt;/math&gt; apples, so we can subtract &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;r&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;y&lt;/math&gt;.<br /> &lt;cmath&gt;\implies (k - 2) + (r - 2) + (y - 2) = 18&lt;/cmath&gt;Let &lt;math&gt;k' = k - 2&lt;/math&gt;, let &lt;math&gt;r' = r - 2&lt;/math&gt;, let &lt;math&gt;y' = y - 2&lt;/math&gt;.<br /> &lt;cmath&gt;\implies k' + r' + y' = 18&lt;/cmath&gt;We can allow either of them to equal to &lt;math&gt;0&lt;/math&gt;, hence this can be solved by stars and bars.<br /> <br /> <br /> By Stars and Bars, our answer is just &lt;math&gt;\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First assume that Alice has &lt;math&gt;2&lt;/math&gt; apples. There are &lt;math&gt;19&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;3&lt;/math&gt; apples, there are &lt;math&gt;18&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;4&lt;/math&gt; apples, there are &lt;math&gt;17&lt;/math&gt; ways to split the rest. So the total number of ways to split &lt;math&gt;24&lt;/math&gt; apples between the three friends is equal to &lt;math&gt;19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let's assume that the three of them have &lt;math&gt;x, y, z&lt;/math&gt; apples. Since each of them has to have at least &lt;math&gt;2&lt;/math&gt; apples, we say that &lt;math&gt;a+2=x, b+2=y&lt;/math&gt; and &lt;math&gt;c+2=z&lt;/math&gt;. Thus, &lt;math&gt;a+b+c+6=24 \implies a+b+c=18&lt;/math&gt;, and so by stars and bars, the number of solutions for this is &lt;math&gt;{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt; - aops5234 <br /> <br /> ==Solution 4==<br /> <br /> Since we have to give each of the &lt;math&gt;3&lt;/math&gt; friends at least &lt;math&gt;2&lt;/math&gt; apples, we need to spend a total of &lt;math&gt;2+2+2=6&lt;/math&gt; apples to solve the restriction. Now we have &lt;math&gt;24-6=18&lt;/math&gt; apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have &lt;math&gt;18&lt;/math&gt; stones and &lt;math&gt;2&lt;/math&gt; sticks, which have a total of &lt;math&gt;\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}&lt;/math&gt; ways to arrange. <br /> <br /> ~by sakshamsethi<br /> <br /> ==Solution 5==<br /> <br /> Equivalently, we split &lt;math&gt;21&lt;/math&gt; apples among &lt;math&gt;3&lt;/math&gt; friends with each having at least &lt;math&gt;1&lt;/math&gt; apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put &lt;math&gt;2&lt;/math&gt; sticks. We have &lt;math&gt; \binom{20}{2} = 190&lt;/math&gt; different ways to arrange the two sticks. So, there are &lt;math&gt;\boxed{190}&lt;/math&gt; ways to split the apples among them.<br /> <br /> ~by Dolphindesigner<br /> <br /> <br /> <br /> ==Video Solutions==<br /> <br /> https://www.youtube.com/channel/UCFoUjg2IndpAryBtHbfOETg?view_as=subscriber<br /> <br /> https://www.youtube.com/watch?v=wJ7uvypbB28<br /> <br /> https://www.youtube.com/watch?v=2dBUklyUaNI<br /> <br /> https://www.youtube.com/watch?v=EJzSOPXULBc<br /> <br /> https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=3qp0wTq-LI0&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=7 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=139083 2019 AMC 8 Problems/Problem 25 2020-12-06T01:13:21Z <p>StellarG: /* Best Video Solution */</p> <hr /> <div>==Problem 25==<br /> Alice has &lt;math&gt;24&lt;/math&gt; apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 1==<br /> We use [[stars and bars]]. Let Alice get &lt;math&gt;k&lt;/math&gt; apples, let Becky get &lt;math&gt;r&lt;/math&gt; apples, let Chris get &lt;math&gt;y&lt;/math&gt; apples.<br /> &lt;cmath&gt;\implies k + r + y = 24&lt;/cmath&gt;We can manipulate this into an equation which can be solved using stars and bars.<br /> <br /> All of them get at least &lt;math&gt;2&lt;/math&gt; apples, so we can subtract &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;r&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;y&lt;/math&gt;.<br /> &lt;cmath&gt;\implies (k - 2) + (r - 2) + (y - 2) = 18&lt;/cmath&gt;Let &lt;math&gt;k' = k - 2&lt;/math&gt;, let &lt;math&gt;r' = r - 2&lt;/math&gt;, let &lt;math&gt;y' = y - 2&lt;/math&gt;.<br /> &lt;cmath&gt;\implies k' + r' + y' = 18&lt;/cmath&gt;We can allow either of them to equal to &lt;math&gt;0&lt;/math&gt;, hence this can be solved by stars and bars.<br /> <br /> <br /> By Stars and Bars, our answer is just &lt;math&gt;\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> First assume that Alice has &lt;math&gt;2&lt;/math&gt; apples. There are &lt;math&gt;19&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;3&lt;/math&gt; apples, there are &lt;math&gt;18&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;4&lt;/math&gt; apples, there are &lt;math&gt;17&lt;/math&gt; ways to split the rest. So the total number of ways to split &lt;math&gt;24&lt;/math&gt; apples between the three friends is equal to &lt;math&gt;19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let's assume that the three of them have &lt;math&gt;x, y, z&lt;/math&gt; apples. Since each of them has to have at least &lt;math&gt;2&lt;/math&gt; apples, we say that &lt;math&gt;a+2=x, b+2=y&lt;/math&gt; and &lt;math&gt;c+2=z&lt;/math&gt;. Thus, &lt;math&gt;a+b+c+6=24 \implies a+b+c=18&lt;/math&gt;, and so by stars and bars, the number of solutions for this is &lt;math&gt;{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt; - aops5234 <br /> <br /> ==Solution 4==<br /> <br /> Since we have to give each of the &lt;math&gt;3&lt;/math&gt; friends at least &lt;math&gt;2&lt;/math&gt; apples, we need to spend a total of &lt;math&gt;2+2+2=6&lt;/math&gt; apples to solve the restriction. Now we have &lt;math&gt;24-6=18&lt;/math&gt; apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have &lt;math&gt;18&lt;/math&gt; stones and &lt;math&gt;2&lt;/math&gt; sticks, which have a total of &lt;math&gt;\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}&lt;/math&gt; ways to arrange. <br /> <br /> ~by sakshamsethi<br /> <br /> ==Solution 5==<br /> <br /> Equivalently, we split &lt;math&gt;21&lt;/math&gt; apples among &lt;math&gt;3&lt;/math&gt; friends with each having at least &lt;math&gt;1&lt;/math&gt; apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put &lt;math&gt;2&lt;/math&gt; sticks. We have &lt;math&gt; \binom{20}{2} = 190&lt;/math&gt; different ways to arrange the two sticks. So, there are &lt;math&gt;\boxed{190}&lt;/math&gt; ways to split the apples among them.<br /> <br /> ~by Dolphindesigner<br /> <br /> <br /> <br /> ==Video Solutions==<br /> <br /> https://www.youtube.com/channel/UCFoUjg2IndpAryBtHbfOETg?view_as=subscriber<br /> <br /> https://www.youtube.com/watch?v=wJ7uvypbB28<br /> <br /> https://www.youtube.com/watch?v=2dBUklyUaNI<br /> <br /> https://www.youtube.com/watch?v=EJzSOPXULBc<br /> <br /> https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=3qp0wTq-LI0&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=7 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Videos explaining solution==<br /> <br /> https://www.youtube.com/watch?v=wJ7uvypbB28<br /> <br /> https://www.youtube.com/watch?v=2dBUklyUaNI<br /> <br /> https://www.youtube.com/watch?v=EJzSOPXULBc<br /> <br /> https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=3qp0wTq-LI0&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=7 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139082 2019 AMC 8 Problems/Problem 24 2020-12-06T01:12:02Z <p>StellarG: /* Solution 20(Straightforward Solution) */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 16 (Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139081 2019 AMC 8 Problems/Problem 24 2020-12-06T01:11:15Z <p>StellarG: /* Solution 19 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139080 2019 AMC 8 Problems/Problem 24 2020-12-06T01:10:49Z <p>StellarG: /* Solution 18 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139079 2019 AMC 8 Problems/Problem 24 2020-12-06T01:10:27Z <p>StellarG: /* Solution 17 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139078 2019 AMC 8 Problems/Problem 24 2020-12-06T01:10:08Z <p>StellarG: /* Solution 16 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 17==<br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx<br /> <br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139077 2019 AMC 8 Problems/Problem 24 2020-12-06T01:09:51Z <p>StellarG: /* Video Solutions */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br /> <br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx <br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25 <br /> <br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==Solution 16==<br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4<br /> - Happytwin (Another video solution)<br /> <br /> ==Solution 17==<br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx<br /> <br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139076 2019 AMC 8 Problems/Problem 24 2020-12-06T01:08:30Z <p>StellarG: /* Video Solutions */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> <br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> ==Solution 16==<br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4<br /> - Happytwin (Another video solution)<br /> <br /> ==Solution 17==<br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx<br /> <br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=139075 2019 AMC 8 Problems/Problem 24 2020-12-06T01:07:52Z <p>StellarG: /* Solution 15 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;\text{Area of Triangle XYZ}&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> &lt;math&gt;AD : DC = 1:2&lt;/math&gt;, so &lt;math&gt;ADB&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt; and &lt;math&gt;CDB&lt;/math&gt; has area &lt;math&gt;240&lt;/math&gt;. &lt;math&gt;BE = ED&lt;/math&gt; so the area of &lt;math&gt;ABE&lt;/math&gt; is equal to the area of &lt;math&gt;ADE = 60&lt;/math&gt;.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Video Solutions==<br /> Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br /> https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br /> <br /> ==Solution 16==<br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4<br /> - Happytwin (Another video solution)<br /> <br /> ==Solution 17==<br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx<br /> <br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> ==Solution 20(Straightforward Solution)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> Since &lt;math&gt;AD:DC=1:2&lt;/math&gt; thus &lt;math&gt;\triangle ABD=\frac{1}{3} \cdot 360 = 120.&lt;/math&gt;<br /> Similarly, &lt;math&gt;\triangle DBC = \frac{2}{3} \cdot 360 = 240.&lt;/math&gt;<br /> Now, since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt;, &lt;math&gt;\triangle ABE = \triangle AED = 120 \div 2 = 60.&lt;/math&gt;<br /> We can use the fact that &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD&lt;/math&gt; even further. Connect lines &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;\triangle BEC&lt;/math&gt; and &lt;math&gt;\triangle DEC&lt;/math&gt; share 2 sides. We know that &lt;math&gt;\triangle BEC=\triangle DEC=240 \div 2 = 120&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is a midpoint of &lt;math&gt;BD.&lt;/math&gt;<br /> Let's label &lt;math&gt;\triangle BEF&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. We know that &lt;math&gt;\triangle EFC&lt;/math&gt; is &lt;math&gt;120-x&lt;/math&gt; since &lt;math&gt;\triangle BEC = 120.&lt;/math&gt;<br /> Note that with this information now, we can deduct more things that are needed to finish the solution.<br /> Note that &lt;math&gt;\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.&lt;/math&gt; because of triangles &lt;math&gt;EBF, ABE, AEC,&lt;/math&gt; and &lt;math&gt;EFC.&lt;/math&gt; We want to find &lt;math&gt;x.&lt;/math&gt;<br /> This is a simple equation, and solving we get &lt;math&gt;x=\boxed{\textbf{(B)}30}.&lt;/math&gt;<br /> <br /> By mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=139074 2019 AMC 8 Problems/Problem 23 2020-12-06T01:06:24Z <p>StellarG: /* Video explaining solution */</p> <hr /> <div>==Problem 23==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points. What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\frac{\text{total points}}{4}&lt;/math&gt; and &lt;math&gt;\frac{2(\text{total points})}{7}&lt;/math&gt; are integers, we have &lt;math&gt;28 | \text{total points}&lt;/math&gt;. We see that the number of points scored by the other team members is less than or equal to &lt;math&gt;14&lt;/math&gt; and greater than or equal to &lt;math&gt;0&lt;/math&gt;. We let the total number of points be &lt;math&gt;t&lt;/math&gt; and the total number of points scored by the other team members be &lt;math&gt;x&lt;/math&gt;, which means that &lt;math&gt;\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14&lt;/math&gt;, which means &lt;math&gt;15 \le \frac{13t}{28} \le 29&lt;/math&gt;. The only value of &lt;math&gt;t&lt;/math&gt; that satisfies all conditions listed is &lt;math&gt;56&lt;/math&gt;, so &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. - juliankuangiccio<br /> <br /> ==Solution 2==<br /> Starting from the above equation &lt;math&gt;\frac{t}{4}+\frac{2t}{7} + 15 + x = t&lt;/math&gt; where &lt;math&gt;t&lt;/math&gt; is the total number of points scored and &lt;math&gt;x\le 14&lt;/math&gt; is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation &lt;math&gt;x+15 = \frac{13}{28}t&lt;/math&gt;, or &lt;math&gt;28x+28\cdot 15=13t&lt;/math&gt;. Since &lt;math&gt;t&lt;/math&gt; is necessarily divisible by 28, let &lt;math&gt;t=28u&lt;/math&gt; where &lt;math&gt;u \ge 0&lt;/math&gt; and divide by 28 to obtain &lt;math&gt;x + 15 = 13u&lt;/math&gt;. Then it is easy to see &lt;math&gt;u=2&lt;/math&gt; (&lt;math&gt;t=56&lt;/math&gt;) is the only candidate, giving &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. -scrabbler94<br /> <br /> ==Solution 3==<br /> We first start by setting the total number of points as &lt;math&gt;28&lt;/math&gt;, since &lt;math&gt;\text{LCM}(4,7) = 28&lt;/math&gt;. However, we see that this does not work since we surpass the number of points just with the information given (&lt;math&gt;28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30&lt;/math&gt; &lt;math&gt;(&gt; 28)&lt;/math&gt; ). Next, we can see that the total number of points scored is &lt;math&gt;56&lt;/math&gt; as, if it is more than or equal to &lt;math&gt;84&lt;/math&gt;, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: &lt;math&gt;56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45&lt;/math&gt;, and thus, the other seven players would have scored a total of &lt;math&gt;56-45 = \boxed{\textbf{(B)} 11}&lt;/math&gt; (We see that this works since we could have &lt;math&gt;4&lt;/math&gt; of them score &lt;math&gt;2&lt;/math&gt; points, and the other &lt;math&gt;3&lt;/math&gt; of them score &lt;math&gt;1&lt;/math&gt; point) -aops5234 -Edited by [[User: Penguin_Spellcaster|Penguin_Spellcaster]]<br /> <br /> ==Solution 4 — Modular Arithmetic ==<br /> <br /> Adding together Alexa's and Brittany's fractions, we get &lt;math&gt;\frac{15}{28}&lt;/math&gt; as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: &lt;math&gt;\frac{15x}{28x}&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; is the common ratio. Let &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have &lt;math&gt;\frac{13x}{28x} = 15 + y + 2z + 0w.&lt;/math&gt; We want all of our variables to be integers. Thus, we want &lt;math&gt;15 + y + 2z = 0 \pmod {13}.&lt;/math&gt; Simplifying, &lt;math&gt;y+2z = 11 \pmod {13}.&lt;/math&gt; The only possible value, as this integer sum has to be less than &lt;math&gt;7 \cdot 2 + 1 = 15,&lt;/math&gt; must be 11. Therefore &lt;math&gt;y+2z = 11,&lt;/math&gt; and the answer is &lt;math&gt;\boxed{ \textbf{(B) 11}}&lt;/math&gt; - ab2024<br /> <br /> ==Video explaining solution== <br /> <br /> Associated video - https://www.youtube.com/watch?v=jE-7Se7ay1c<br /> <br /> https://www.youtube.com/watch?v=3Mae_6qFxoU&amp;t=204s ~ hi_im_bob<br /> <br /> https://youtu.be/wsYCn2FqZJE<br /> <br /> https://www.youtube.com/watch?v=fKjmw_zzCUU<br /> <br /> https://www.youtube.com/watch?v=o2mcnLOVFBA&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=5 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=138130 2019 AMC 8 Problems/Problem 22 2020-11-22T21:44:31Z <p>StellarG: /* Video explaining solution */</p> <hr /> <div>==Problem 22==<br /> A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose the fraction of discount is &lt;math&gt;x&lt;/math&gt;. That means &lt;math&gt;(1-x)(1+x)=0.84&lt;/math&gt;; so &lt;math&gt;1-x^{2}=0.84&lt;/math&gt;, and &lt;math&gt;(x^{2})=0.16&lt;/math&gt;, obtaining &lt;math&gt;x=0.4&lt;/math&gt;. Therefore, the price was increased and decreased by &lt;math&gt;40&lt;/math&gt;%, or &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Answer options)==<br /> We can try out every option and see which one works out. By this method, we get &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;. ~avamarora<br /> <br /> ==Solution 3==<br /> Let x be the discount. We can also work in reverse such as (&lt;math&gt;84&lt;/math&gt;)&lt;math&gt;(\frac{100}{100-x})&lt;/math&gt;&lt;math&gt;(\frac{100}{100+x})&lt;/math&gt; = &lt;math&gt;100&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;8400&lt;/math&gt; = &lt;math&gt;(100+x)(100-x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; gives us &lt;math&gt;x = 40, -40&lt;/math&gt;. But &lt;math&gt;x&lt;/math&gt; has to be positive. Thus &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;40&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> <br /> ==Solution 4 ~ using the answer choices==<br /> <br /> Let our original cost be &lt;math&gt;\$100.&lt;/math&gt; We are looking for a result of &lt;math&gt;\$ 84,&lt;/math&gt; then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try &lt;math&gt;\boxed{40\%}&lt;/math&gt;, and we have the answer; it worked.<br /> <br /> ==Video explaining solution== <br /> <br /> Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br /> <br /> https://youtu.be/gX_l0PGsQao<br /> <br /> https://www.youtube.com/watch?v=_TheVi-6LWE<br /> <br /> https://www.youtube.com/watch?v=RcBDdB35Whk&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=4 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_21&diff=138129 2019 AMC 8 Problems/Problem 21 2020-11-22T21:43:47Z <p>StellarG: /* Video Solutions */</p> <hr /> <div><br /> ==Problem 21==<br /> What is the area of the triangle formed by the lines &lt;math&gt;y=5&lt;/math&gt;, &lt;math&gt;y=1+x&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First we need to find the coordinates where the graphs intersect. <br /> <br /> We want the points x and y to be the same. Thus, we set &lt;math&gt;5=x+1,&lt;/math&gt; and get &lt;math&gt;x=4.&lt;/math&gt; Plugging this into the equation, &lt;math&gt;y=1-x,&lt;/math&gt; <br /> &lt;math&gt;y=5&lt;/math&gt;, and &lt;math&gt;y=1+x&lt;/math&gt; intersect at &lt;math&gt;(4,5)&lt;/math&gt;, we call this line x.<br /> <br /> Doing the same thing, we get &lt;math&gt;x=-4.&lt;/math&gt; Thus &lt;math&gt;y=5&lt;/math&gt; also.<br /> &lt;math&gt;y=5&lt;/math&gt;, and &lt;math&gt;y=1-x&lt;/math&gt; intersect at &lt;math&gt;(-4,5)&lt;/math&gt;, we call this line y.<br /> <br /> It's apparent the only solution to &lt;math&gt;1-x=1+x&lt;/math&gt; is &lt;math&gt;0.&lt;/math&gt; Thus, &lt;math&gt;y=1.&lt;/math&gt;<br /> &lt;math&gt;y=1-x&lt;/math&gt; and &lt;math&gt;y=1+x&lt;/math&gt; intersect at &lt;math&gt;(0,1)&lt;/math&gt;, we call this line z.<br /> <br /> Using the [[Shoelace Theorem]] we get: &lt;cmath&gt;\left(\frac{(20-4)-(-20+4)}{2}\right)=\frac{32}{2}&lt;/cmath&gt; &lt;math&gt;=&lt;/math&gt; So our answer is &lt;math&gt;\boxed{\textbf{(E)}\ 16.}&lt;/math&gt;<br /> <br /> We might also see that the lines &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are mirror images of each other. This is because, when rewritten, their slopes can be multiplied by &lt;math&gt;-1&lt;/math&gt; to get the other. As the base is horizontal, this is a isosceles triangle with base 8, as the intersection points have distance 8. The height is &lt;math&gt;5-1=4,&lt;/math&gt; so &lt;math&gt;\frac{4\cdot 8}{2} = \boxed{\textbf{(E)} 16.}&lt;/math&gt;<br /> <br /> Warning: Do not use the distance formula for the base then use heron's formula. It will take you half of the time you have left!<br /> <br /> ==Solution 2==<br /> Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get &lt;math&gt;\frac{4\cdot8}{2}&lt;/math&gt; which is equal to &lt;math&gt;\boxed{\textbf{(E)}\ 16}&lt;/math&gt;.<br /> <br /> ==Video Solutions==<br /> Associated Video - https://www.youtube.com/watch?v=ie3tlSNyiaY<br /> <br /> https://www.youtube.com/watch?v=9nlX9VCisQc<br /> <br /> https://www.youtube.com/watch?v=mz3DY1rc5ao<br /> <br /> https://www.youtube.com/watch?v=Z27G0xy5AgA&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=3 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=20|num-a=22}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_20&diff=138128 2019 AMC 8 Problems/Problem 20 2020-11-22T21:42:58Z <p>StellarG: /* Solution 3 */</p> <hr /> <div>==Problem 20==<br /> How many different real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;cmath&gt;(x^{2}-5)^{2}=16?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We have that &lt;math&gt;(x^2-5)^2 = 16&lt;/math&gt; if and only if &lt;math&gt;x^2-5 = \pm 4&lt;/math&gt;. If &lt;math&gt;x^2-5 = 4&lt;/math&gt;, then &lt;math&gt;x^2 = 9 \implies x = \pm 3&lt;/math&gt;, giving 2 solutions. If &lt;math&gt;x^2-5 = -4&lt;/math&gt;, then &lt;math&gt;x^2 = 1 \implies x = \pm 1&lt;/math&gt;, giving 2 more solutions. All four of these solutions work, so the answer is &lt;math&gt;\boxed{\textbf{(D)} 4}&lt;/math&gt;. Further, the equation is a quartic in &lt;math&gt;x&lt;/math&gt;, so by the [https://artofproblemsolving.com/wiki/index.php/Fundamental_Theorem_of_Algebra Fundamental Theorem of Algebra], there can be at most four real solutions.<br /> <br /> ==Solution 2==<br /> We can expand &lt;math&gt;(x^2-5)^2&lt;/math&gt; to get &lt;math&gt;x^4-10x^2+25&lt;/math&gt;, so now our equation is &lt;math&gt;x^4-10x^2+25=16&lt;/math&gt;. Subtracting &lt;math&gt;16&lt;/math&gt; from both sides gives us &lt;math&gt;x^4-10x^2+9=0&lt;/math&gt;. Now, we can factor the left hand side to get &lt;math&gt;(x^2-9)(x^2-1)=0&lt;/math&gt;. If &lt;math&gt;x^2-9&lt;/math&gt; and/or &lt;math&gt;x^2-1&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt;, then the whole left side will equal &lt;math&gt;0&lt;/math&gt;. Since the solutions can be both positive and negative, we have &lt;math&gt;4&lt;/math&gt; solutions: &lt;math&gt;-3,3,-1,1&lt;/math&gt; (we can find these solutions by setting &lt;math&gt;x^2-9&lt;/math&gt; and &lt;math&gt;x^2-1&lt;/math&gt; equal to &lt;math&gt;0&lt;/math&gt; and solving for &lt;math&gt;x&lt;/math&gt;). So the answer is &lt;math&gt;\boxed{\textbf{(D)} 4}&lt;/math&gt;. <br /> <br /> ~UnstoppableGoddess<br /> <br /> ==Solution 3==<br /> Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw<br /> <br /> https://youtu.be/0AY1klX3gBo<br /> <br /> ==Solution 4==<br /> https://youtu.be/5BXh0JY4klM (Uses a difference of squares &amp; factoring method, different from above solutions)<br /> <br /> ==Solution 5 (Fast)==<br /> https://www.youtube.com/watch?v=44vrsk_CbF8&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=2 ~ MathEx<br /> <br /> ==Solution 6==<br /> Video Solution - https://youtu.be/Lw8fSbX_8FU<br /> <br /> ==Video Solution==<br /> https://youtu.be/V3HxkJhSn08<br /> <br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=19|num-a=21}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_19&diff=134442 2019 AMC 8 Problems/Problem 19 2020-10-01T17:11:59Z <p>StellarG: /* Solution 3 */</p> <hr /> <div>==Problem 19==<br /> In a tournament there are six teams that play each other twice. A team earns &lt;math&gt;3&lt;/math&gt; points for a win, &lt;math&gt;1&lt;/math&gt; point for a draw, and &lt;math&gt;0&lt;/math&gt; points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?<br /> <br /> &lt;math&gt;\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> After fully understanding the problem, we immediately know that the three top teams, say team &lt;math&gt;A&lt;/math&gt;, team &lt;math&gt;B&lt;/math&gt;, and team &lt;math&gt;C&lt;/math&gt;, must beat the other three teams &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;. Therefore, &lt;math&gt;A&lt;/math&gt;,&lt;math&gt;B&lt;/math&gt;,&lt;math&gt;C&lt;/math&gt; must each obtain &lt;math&gt;(3+3+3)=9&lt;/math&gt; points. However, they play against each team twice, for a total of &lt;math&gt;18&lt;/math&gt; points against &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;. For games between &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, we have 2 cases. In both cases, there is an equality of points between &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Case 1: A team ties the two other teams. For a tie, we have 1 point, so we have &lt;math&gt;(1+1)*2=4&lt;/math&gt; points (they play twice). Therefore, this case brings a total of &lt;math&gt;4+18=22&lt;/math&gt; points.<br /> <br /> Case 2: A team beats one team while losing to another. This gives equality, as each team wins once and loses once as well. For a win, we have &lt;math&gt;3&lt;/math&gt; points, so a team gets &lt;math&gt;3\times2=6&lt;/math&gt; points if they each win a game and lose a game. This case brings a total of &lt;math&gt;18+6=24&lt;/math&gt; points. <br /> <br /> Therefore, we use Case 2 since it brings the greater amount of points, or &lt;math&gt;\boxed{24}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> --------------------------<br /> Note that case 2 can be easily seen to be better as follows. Let &lt;math&gt;x_A&lt;/math&gt; be the number of points &lt;math&gt;A&lt;/math&gt; gets, &lt;math&gt;x_B&lt;/math&gt; be the number of points &lt;math&gt;B&lt;/math&gt; gets, and &lt;math&gt;x_C&lt;/math&gt; be the number of points &lt;math&gt;C&lt;/math&gt; gets. Since &lt;math&gt;x_A = x_B = x_C&lt;/math&gt;, to maximize &lt;math&gt;x_A&lt;/math&gt;, we can just maximize &lt;math&gt;x_A + x_B + x_C&lt;/math&gt;. But in each match, if one team wins then the total sum increases by &lt;math&gt;3&lt;/math&gt; points, whereas if they tie, the total sum increases by &lt;math&gt;2&lt;/math&gt; points. So it is best if there are the fewest ties possible.<br /> <br /> ==Solution 2==<br /> <br /> (1st match(3) + 2nd match(1)) * number of teams(6) = 24, &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> Explanation: So after reading the problem we see that there are 6 teams and each team versus each other twice. This means one of the two matches has to be a win, so 3 points so far. Now if we say that the team won again and make it 6 points, that would mean that team would be dominating the leader-board and the problem says that all the top 3 people have the same score. So that means the maximum amount of points we could get is 1 so that each team gets the same amount of matches won &amp; drawn so that adds up to 4. 4 * the number of teams(6) = 24 so the answer is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> ~MRLUIGI<br /> <br /> <br /> ==Solution 3==<br /> <br /> <br /> We can name the top three teams as &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. We can see that &lt;math&gt;A=B=C&lt;/math&gt;, because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: &lt;math&gt;AB, BC,&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; come twice. In order to even out the scores and get the maximum score, we can say that in match &lt;math&gt;AB, A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each win once out of the two games that they play. We can say the same thing for &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;. This tells us that each team &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win and lose twice. This gives each team a total of 3 + 3 + 0 + 0 = 6 points. Now, we need to include the other three teams. We can label these teams as &lt;math&gt;D, E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;. We can write down every match that &lt;math&gt;A, B,&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; plays in that we haven't counted yet: &lt;math&gt;AD, AD, AE, AE, AF, AF, BD, BD, BE, BE, BF, BF, CD, CD, CE, CE, CF,&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt;. We can say &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win each of these in order to obtain the maximum score that &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; can have. If &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; win all six of their matches, &lt;math&gt;A, B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; will have a score of &lt;math&gt;18&lt;/math&gt;. &lt;math&gt;18 + 6&lt;/math&gt; results in a maximum score of &lt;math&gt;\boxed{24}&lt;/math&gt;. This tells us that the correct answer choice is &lt;math&gt;\boxed{C}&lt;/math&gt;.<br /> <br /> ~Champion1234<br /> <br /> <br /> == Video Solutions ==<br /> <br /> Associated Video - https://youtu.be/s0O3_uXZrOI<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_18&diff=134441 2019 AMC 8 Problems/Problem 18 2020-10-01T17:09:33Z <p>StellarG: /* Video Solution */</p> <hr /> <div>==Problem 18==<br /> The faces of each of two fair dice are numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;8&lt;/math&gt;. When the two dice are tossed, what is the probability that their sum will be an even number?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{4}{9}\qquad\textbf{(B) }\frac{1}{2}\qquad\textbf{(C) }\frac{5}{9}\qquad\textbf{(D) }\frac{3}{5}\qquad\textbf{(E) }\frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The approach to this problem:<br /> There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order) . We can solve for how many ways the &lt;math&gt;2&lt;/math&gt; numbers add up to an odd number and subtract the answer from &lt;math&gt;1&lt;/math&gt;. <br /> <br /> How to solve the problem:<br /> The probability of getting an odd number first is &lt;math&gt;\frac{4}{6}=\frac{2}{3}&lt;/math&gt;. In order to make the sum odd, we must select an even number next. The probability of getting an even number is &lt;math&gt;\frac{2}{6}=\frac{1}{3}&lt;/math&gt;. Now we multiply the two fractions: &lt;math&gt;\frac{2}{3}\times\frac{1}{3}=2/9&lt;/math&gt;. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except backward and by the Communitive Property of Multiplication, the equations are it does not matter is the equation is backward or not. Thus we do &lt;math&gt;\frac{2}{9}\times2=\frac{4}{9}&lt;/math&gt;. This is the probability of getting an odd number. In order to get the probability of getting an even number we do &lt;math&gt;1-\frac{4}{9}=\boxed{(\textbf{C})\frac{5}{9}}&lt;/math&gt;<br /> <br /> - ViratKohli2018 (VK18)<br /> <br /> ==Solution 2==<br /> We have a &lt;math&gt;2&lt;/math&gt; die with &lt;math&gt;2&lt;/math&gt; evens and &lt;math&gt;4&lt;/math&gt; odds on both dies. For the sum to be even, the rolls must consist of &lt;math&gt;2&lt;/math&gt; odds or &lt;math&gt;2&lt;/math&gt; evens. <br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; odds (Case &lt;math&gt;1&lt;/math&gt;): The total number of ways to roll &lt;math&gt;2&lt;/math&gt; odds is &lt;math&gt;4*4=16&lt;/math&gt;, as there are &lt;math&gt;4&lt;/math&gt; choices for the first odd on the first roll and &lt;math&gt;4&lt;/math&gt; choices for the second odd on the second roll.<br /> <br /> Ways to roll &lt;math&gt;2&lt;/math&gt; evens (Case &lt;math&gt;2&lt;/math&gt;): Similarly, we have &lt;math&gt;2*2=4&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; evens.<br /> <br /> Totally, we have &lt;math&gt;6*6=36&lt;/math&gt; ways to roll &lt;math&gt;2&lt;/math&gt; dies.<br /> <br /> Therefore the answer is &lt;math&gt;\frac{16+4}{36}=\frac{20}{36}=\frac{5}{9}&lt;/math&gt;, or &lt;math&gt;\framebox{C}&lt;/math&gt;.<br /> <br /> ~A1337h4x0r<br /> <br /> ==Solution 3 (Complementary Counting)==<br /> We count the ways to get an odd. If the sum is odd, then we must have an even and an odd. The probability of an even is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the probability of an odd is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. We have to multiply by &lt;math&gt;2!&lt;/math&gt; because the even and odd can be in any order. This gets us &lt;math&gt;\frac{4}{9}&lt;/math&gt;, so the answer is &lt;math&gt;1 - \frac{4}{9} = \frac{5}{9} = \boxed{(\textbf{C})}&lt;/math&gt;. - juliankuang<br /> <br /> ==Solution 4==<br /> To get an even, you must get either 2 odds or 2 evens. The probability of getting 2 odds is &lt;math&gt;\frac{4}{6} * \frac{4}{6}&lt;/math&gt;. The probability of getting 2 evens is &lt;math&gt;\frac{2}{6} * \frac{2}{6}&lt;/math&gt;. If you add them together, you get &lt;math&gt;\frac{16}{36} + \frac{4}{36}&lt;/math&gt; = &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~heeeeeeeeheeeee<br /> <br /> ==Solution 5 (Casework)==<br /> To get an even number, we must either have two odds or two evens. We will solve this through casework. The probability of rolling a 1 is 1/6, and the probability of rolling another odd number after this is 4/6=2/3, so the probability of getting a sum of an even number is (1/6)(2/3)=1/9. The probability of rolling a 2 is 1/6, and the probability of rolling another even number after this is 2/6=1/3, so the probability of rolling a sum of an even number is (1/6)(1/3)=1/18. Now, notice that the probability of getting an even sum with two odd numbers is identical for all odd numbers. This is because the probability of rolling the desired number is 1/6, and the probability of rolling a second odd number is constant, being 2/3. Since there are four cases that involve odd numbers, the probability of getting an even sum with two odd numbers is (4)(1/9)=4/9. Using this same logic, realize that the probability of getting an even number is identical for all even numbers, so the probability of getting an even sum with only even numbers is (2)(1/18)=1/9. Adding these two up, we get our desired &lt;math&gt;\boxed{(\textbf{C}) \frac{5}{9}}&lt;/math&gt;.~binderclips1<br /> <br /> ==Video Solution==<br /> <br /> Associated video - https://www.youtube.com/watch?v=EoBZy_WYWEw<br /> <br /> https://www.youtube.com/watch?v=H52AqAl4nt4&amp;t=2s ~ MathEx<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_17&diff=134440 2019 AMC 8 Problems/Problem 17 2020-10-01T17:08:59Z <p>StellarG: /* Video Solution */</p> <hr /> <div>==Problem 17==<br /> What is the value of the product <br /> &lt;cmath&gt;\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50&lt;/math&gt;<br /> <br /> ==Solution 1(Telescoping)==<br /> We rewrite: &lt;cmath&gt;\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}&lt;/cmath&gt;<br /> <br /> The middle terms cancel, leaving us with<br /> <br /> &lt;cmath&gt;\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}&lt;/cmath&gt;<br /> <br /> ~ hpotter0104<br /> <br /> ==Solution 2==<br /> If you calculate the first few values of the equation, all of the values tend to &lt;math&gt;\frac{1}{2}&lt;/math&gt;, but are not equal to it. The answer closest to &lt;math&gt;\frac{1}{2}&lt;/math&gt; but not equal to it is &lt;math&gt;\boxed{\textbf{(B)}\frac{50}{99}}&lt;/math&gt;.~hpotter0104<br /> <br /> ==Solution 3==<br /> Rewriting the numerator and the denominator, we get &lt;math&gt;\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}&lt;/math&gt;. We can simplify by canceling 99! on both sides, leaving us with: &lt;math&gt;\frac{100 \cdot 98!}{2 \cdot 99!}&lt;/math&gt; We rewrite &lt;math&gt;99!&lt;/math&gt; as &lt;math&gt;99 \cdot 98!&lt;/math&gt; and cancel &lt;math&gt;98!&lt;/math&gt;, which gets &lt;math&gt;\boxed{\frac{50}{99}}&lt;/math&gt;. Answer B.<br /> <br /> ==Video Solution==<br /> <br /> Associated video - https://www.youtube.com/watch?v=yPQmvyVyvaM<br /> <br /> https://www.youtube.com/watch?v=ffHl1dAjs7g&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=1 ~ MathEx<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_16&diff=134439 2019 AMC 8 Problems/Problem 16 2020-10-01T17:08:24Z <p>StellarG: /* Video Solution */</p> <hr /> <div>==Problem 16==<br /> Qiang drives &lt;math&gt;15&lt;/math&gt; miles at an average speed of &lt;math&gt;30&lt;/math&gt; miles per hour. How many additional miles will he have to drive at &lt;math&gt;55&lt;/math&gt; miles per hour to average &lt;math&gt;50&lt;/math&gt; miles per hour for the entire trip?<br /> <br /> &lt;math&gt;\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The only option that is easily divisible by &lt;math&gt;55&lt;/math&gt; is &lt;math&gt;110&lt;/math&gt;. Which gives 2 hours of travel. And by the formula &lt;math&gt;\frac{15}{30} + \frac{110}{55} = \frac{5}{2}&lt;/math&gt;<br /> <br /> And &lt;math&gt;\text{Average Speed}&lt;/math&gt; = &lt;math&gt;\frac{\text{Total Distance}}{\text{Total Time}}&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{125}{50} = \frac{5}{2}&lt;/math&gt;<br /> <br /> Both are equal and thus our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Note that the average speed is simply the total distance over the total time. Let the number of additional miles he has to drive be &lt;math&gt;x.&lt;/math&gt; Therefore, the total distance is &lt;math&gt;15+x&lt;/math&gt; and the total time (in hours) is &lt;cmath&gt;\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.&lt;/cmath&gt; We can set up the following equation: &lt;cmath&gt;\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.&lt;/cmath&gt; Simplifying the equation, we get &lt;cmath&gt;15+x=25+\frac{10x}{11}.&lt;/cmath&gt; Solving the equation yields &lt;math&gt;x=110,&lt;/math&gt; so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}&lt;/math&gt;. <br /> <br /> ~twinemma<br /> <br /> ==Solution 3==<br /> If he travels &lt;math&gt;15&lt;/math&gt; miles at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour, he travels for 30 min. Average rate is total distance over total time so &lt;math&gt;(15+d)/(0.5 + t) = 50&lt;/math&gt;, where d is the distance left to travel and t is the time to travel that distance. solve for &lt;math&gt;d&lt;/math&gt; to get &lt;math&gt;d = 10+50t&lt;/math&gt;. you also know that he has to travel &lt;math&gt;55&lt;/math&gt; miles per hour for some time, so &lt;math&gt;d=55t&lt;/math&gt; plug that in for d to get &lt;math&gt;55t = 10+50t&lt;/math&gt; and &lt;math&gt;t=2&lt;/math&gt; and since &lt;math&gt;d=55t&lt;/math&gt;, &lt;math&gt;d = 2\cdot55 =110&lt;/math&gt; the answer is &lt;math&gt;\boxed{\textbf{(D)}\ 110}&lt;/math&gt;.<br /> <br /> -goldenn<br /> <br /> ==Video Solution==<br /> <br /> Associated Video - https://www.youtube.com/watch?v=OC1KdFeZFeE<br /> <br /> https://youtu.be/5K1AgeZ8rUQ - happytwin<br /> <br /> Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_15&diff=128507 2019 AMC 8 Problems/Problem 15 2020-07-18T00:45:45Z <p>StellarG: /* Video Solution */</p> <hr /> <div>==Problem 15==<br /> On a beach &lt;math&gt;50&lt;/math&gt; people are wearing sunglasses and &lt;math&gt;35&lt;/math&gt; people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is is also wearing sunglasses is &lt;math&gt;\frac{2}{5}&lt;/math&gt;. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?<br /> &lt;math&gt;\textbf{(A) }\frac{14}{85}\qquad\textbf{(B) }\frac{7}{25}\qquad\textbf{(C) }\frac{2}{5}\qquad\textbf{(D) }\frac{4}{7}\qquad\textbf{(E) }\frac{7}{10}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The number of people wearing caps and sunglasses is <br /> &lt;math&gt;\frac{2}{5}\cdot35=14&lt;/math&gt;. So then 14 people out of the 50 people wearing sunglasses also have caps. <br /> &lt;math&gt;\frac{14}{50}=\boxed{\textbf{(B)}\frac{7}{25}}&lt;/math&gt;<br /> <br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=gKlYlAiBzrs ~ MathEx<br /> <br /> Another video - https://www.youtube.com/watch?v=afMsUqER13c<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=14|num-a=16}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=128504 2019 AMC 8 Problems/Problem 14 2020-07-18T00:44:51Z <p>StellarG: /* Solution 5 */</p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;Wednesday&lt;/math&gt; she redeems her next coupon on &lt;math&gt;Saturday&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be &quot;used&quot; on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her eleventh coupon, of which she only has ten. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> == Solution 6 ==<br /> Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&diff=128502 2019 AMC 8 Problems/Problem 13 2020-07-18T00:43:12Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 13==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that the only positive 2-digit palindromes are multiples of 11, namely &lt;math&gt;11, 22, \ldots, 99&lt;/math&gt;. Since &lt;math&gt;N&lt;/math&gt; is the sum of 2-digit palindromes, &lt;math&gt;N&lt;/math&gt; is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so &lt;math&gt;N=110&lt;/math&gt; is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as &lt;math&gt;110=77+22+11&lt;/math&gt;. Then &lt;math&gt;N = 110&lt;/math&gt;, and the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;1+1+0 = \boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Associated video - https://www.youtube.com/watch?v=bOnNFeZs7S8<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=12|num-a=14}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=128501 2019 AMC 8 Problems/Problem 12 2020-07-18T00:33:48Z <p>StellarG: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]]<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;B&lt;/math&gt; is on the top, and &lt;math&gt;R&lt;/math&gt; is on the side, and &lt;math&gt;G&lt;/math&gt; is on the right side. That means that (image &lt;math&gt;2&lt;/math&gt;) &lt;math&gt;W&lt;/math&gt; is on the left side. From the third image, you know that &lt;math&gt;P&lt;/math&gt; must be on the bottom since &lt;math&gt;G&lt;/math&gt; is sideways. That leaves us with the back, so the back must be &lt;math&gt;A&lt;/math&gt;. The front is opposite of the back, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.~heeeeeeeheeeee<br /> <br /> ==Solution 2==<br /> Looking closely we can see that all faces are connected with &lt;math&gt;R&lt;/math&gt; except for &lt;math&gt;A&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> <br /> ~phoenixfire<br /> <br /> ==Solution 3==<br /> Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM<br /> <br /> ==Note==<br /> {{AMC8 box|year=2019|num-b=11|num-a=13}}<br /> Only two of the cubes are required to solve the problem.</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=128499 2019 AMC 8 Problems/Problem 12 2020-07-18T00:33:27Z <p>StellarG: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]]<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;B&lt;/math&gt; is on the top, and &lt;math&gt;R&lt;/math&gt; is on the side, and &lt;math&gt;G&lt;/math&gt; is on the right side. That means that (image &lt;math&gt;2&lt;/math&gt;) &lt;math&gt;W&lt;/math&gt; is on the left side. From the third image, you know that &lt;math&gt;P&lt;/math&gt; must be on the bottom since &lt;math&gt;G&lt;/math&gt; is sideways. That leaves us with the back, so the back must be &lt;math&gt;A&lt;/math&gt;. The front is opposite of the back, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.~heeeeeeeheeeee<br /> <br /> ==Solution 2==<br /> Looking closely we can see that all faces are connected with &lt;math&gt;R&lt;/math&gt; except for &lt;math&gt;A&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> <br /> ~phoenixfire<br /> <br /> ==Solution 4==<br /> Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM<br /> <br /> ==Note==<br /> {{AMC8 box|year=2019|num-b=11|num-a=13}}<br /> Only two of the cubes are required to solve the problem.</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=128498 2019 AMC 8 Problems/Problem 11 2020-07-18T00:33:00Z <p>StellarG: /* Solution 4 */</p> <hr /> <div>==Problem 11==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of students taking both a math and a foreign language class.<br /> <br /> By P-I-E, we get &lt;math&gt;70 + 54 - x&lt;/math&gt; = &lt;math&gt;93&lt;/math&gt;. <br /> <br /> Solving gives us &lt;math&gt;x = 31&lt;/math&gt;.<br /> <br /> But we want the number of students taking only a math class.<br /> <br /> Which is &lt;math&gt;70 - 31 = 39&lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 39}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 2==<br /> We have &lt;math&gt;70 + 54 = 124&lt;/math&gt; people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that &lt;math&gt;31&lt;/math&gt; people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get &lt;math&gt;70 - 31 = \boxed{\textbf{D} \, 39}&lt;/math&gt; -fath2012<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> draw(circle((-0.5,0),1));<br /> draw(circle((0.5,0),1));<br /> label(&quot;$\huge{x}$&quot;, (0, 0));<br /> label(&quot;$70-x$&quot;, (-1, 0));<br /> label(&quot;$54-x$&quot;, (1, 0));<br /> &lt;/asy&gt;<br /> <br /> We know that the sum of all three areas is &lt;math&gt;93&lt;/math&gt;<br /> So, we have: <br /> &lt;cmath&gt;93 = 70-x+x+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 70+54-x&lt;/cmath&gt;<br /> &lt;cmath&gt;93 = 124 - x&lt;/cmath&gt;<br /> &lt;cmath&gt;-39=-x&lt;/cmath&gt;<br /> &lt;cmath&gt;x=39&lt;/cmath&gt;<br /> <br /> We are looking for the number of students in only math. This is &lt;math&gt;70-x&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; with &lt;math&gt;31&lt;/math&gt;, our answer is &lt;math&gt;\boxed{39}&lt;/math&gt;.<br /> <br /> -mathnerdnair<br /> <br /> ==Solution 4==<br /> Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&diff=125397 2019 AMC 8 Problems/Problem 12 2020-06-14T16:58:32Z <p>StellarG: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> The faces of a cube are painted in six different colors: red &lt;math&gt;(R)&lt;/math&gt;, white &lt;math&gt;(W)&lt;/math&gt;, green &lt;math&gt;(G)&lt;/math&gt;, brown &lt;math&gt;(B)&lt;/math&gt;, aqua &lt;math&gt;(A)&lt;/math&gt;, and purple &lt;math&gt;(P)&lt;/math&gt;. Three views of the cube are shown below. What is the color of the face opposite the aqua face?<br /> <br /> [[File:2019AMC8Prob12.png]]<br /> <br /> &lt;math&gt;\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;B&lt;/math&gt; is on the top, and &lt;math&gt;R&lt;/math&gt; is on the side, and &lt;math&gt;G&lt;/math&gt; is on the right side. That means that (image &lt;math&gt;2&lt;/math&gt;) &lt;math&gt;W&lt;/math&gt; is on the left side. From the third image, you know that &lt;math&gt;P&lt;/math&gt; must be on the bottom since &lt;math&gt;G&lt;/math&gt; is sideways. That leaves us with the back, so the back must be &lt;math&gt;A&lt;/math&gt;. The front is opposite of the back, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.~heeeeeeeheeeee<br /> <br /> ==Solution 2==<br /> Looking closely we can see that all faces are connected with &lt;math&gt;R&lt;/math&gt; except for &lt;math&gt;A&lt;/math&gt;. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ R}&lt;/math&gt;.<br /> <br /> <br /> ~phoenixfire<br /> <br /> <br /> Video solution: https://www.youtube.com/watch?v=K5vaX_EzjEM<br /> <br /> ==Note==<br /> {{AMC8 box|year=2019|num-b=11|num-a=13}}<br /> Only two of the cubes are required to solve the problem.</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_11&diff=122198 2019 AMC 8 Problems/Problem 11 2020-05-09T16:24:34Z <p>StellarG: /* Solution 2 */</p> <hr /> <div>==Problem 11==<br /> The eighth grade class at Lincoln Middle School has &lt;math&gt;93&lt;/math&gt; students. Each student takes a math class or a foreign language class or both. There are &lt;math&gt;70&lt;/math&gt; eighth graders taking a math class, and there are &lt;math&gt;54&lt;/math&gt; eight graders taking a foreign language class. How many eighth graders take ''only'' a math class and ''not'' a foreign language class?<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }23\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; be the number of students taking both a math and a foreign language class.<br /> <br /> By P-I-E, we get &lt;math&gt;70 + 54 - x&lt;/math&gt; = &lt;math&gt;93&lt;/math&gt;. <br /> <br /> Solving gives us &lt;math&gt;x = 31&lt;/math&gt;.<br /> <br /> But we want the number of students taking only a math class.<br /> <br /> Which is &lt;math&gt;70 - 31 = 39&lt;/math&gt;.<br /> <br /> &lt;math&gt;\boxed{\textbf{(D)}\ 39}&lt;/math&gt;<br /> <br /> ~phoenixfire<br /> <br /> ==Solution 2==<br /> We have &lt;math&gt;70 + 54 = 124&lt;/math&gt; people taking classes. However we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that &lt;math&gt;31&lt;/math&gt; people took the two classes. To find the amount of people who took only math class web subtract the people who didn't take only one math class, so we get &lt;math&gt;70 - 31 = \boxed{\textbf{D} \, 39}&lt;/math&gt; -fath2012<br /> <br /> ==Solution 3==<br /> Associated video - https://www.youtube.com/watch?v=onPaMTO3dSA<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems&diff=118292 2020 AMC 10B Problems 2020-02-21T23:14:08Z <p>StellarG: /* Problem 1 */</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;1 - (-2) - 3 - (-4) - 5 - (-6)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> Editing 2020 AMC 10B Problems/Problem 2 (section)<br /> <br /> ==Problem 2==<br /> <br /> Aaron has &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;1&lt;/math&gt;, and Steph and Val each have &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;2&lt;/math&gt;. What is the total volume of these &lt;math&gt;10&lt;/math&gt; cubes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The acute angles of a right triangle are &lt;math&gt;a^{\circ}&lt;/math&gt; and &lt;math&gt;b^{\circ}&lt;/math&gt;, where &lt;math&gt;a&gt;b&lt;/math&gt; and both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are prime numbers. What is the least possible value of &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Driving along a highway, Megan noticed that her odometer showed &lt;math&gt;15951&lt;/math&gt; (miles). This number is a palindrome-it reads the same forward and backward. Then &lt;math&gt;2&lt;/math&gt; hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this &lt;math&gt;2&lt;/math&gt;-hour period?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many positive even multiples of &lt;math&gt;3&lt;/math&gt; less than &lt;math&gt;2020&lt;/math&gt; are perfect squares?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie in a plane with &lt;math&gt;PQ=8&lt;/math&gt;. How many locations for point &lt;math&gt;R&lt;/math&gt; in this plane are there such that the triangle with vertices &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; is a right triangle with area &lt;math&gt;12&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> How many ordered pairs of integers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the equation &lt;cmath&gt;x^{2020} + y^2 = 2y?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A three-quarter sector of a circle of radius &lt;math&gt;4&lt;/math&gt; inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?<br /> &lt;asy&gt;<br /> <br /> draw(Arc((0,0), 4, 0, 270));<br /> draw((0,-4)--(0,0)--(4,0));<br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Ms.Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> The decimal representation of &lt;cmath&gt;\frac{1}{20^{20}}&lt;/cmath&gt;<br /> consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Andy the Ant lives on a coordinate plane and is currently at &lt;math&gt;(-20, 20)&lt;/math&gt; facing east (that is, in the positive &lt;math&gt;x&lt;/math&gt;-direction). Andy moves &lt;math&gt;1&lt;/math&gt; unit and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. From there, Andy moves &lt;math&gt;2&lt;/math&gt; units (north) and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. He then moves &lt;math&gt;3&lt;/math&gt; units (west) and again turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. Andy continues his progress, increasing his distance each time by &lt;math&gt;1&lt;/math&gt; unit and always turning left. What is the location of the point at which Andy makes the &lt;math&gt;2020&lt;/math&gt;th left turn?<br /> <br /> &lt;math&gt;\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?<br /> <br /> &lt;asy&gt;<br /> real x=sqrt(3);<br /> real y=2sqrt(3);<br /> real z=3.5;<br /> real a=x/2;<br /> real b=0.5;<br /> real c=3a;<br /> pair A, B, C, D, E, F;<br /> A = (1,0);<br /> B = (3,0);<br /> C = (4,x);<br /> D = (3,y);<br /> E = (1,y);<br /> F = (0,x);<br /> <br /> fill(A--B--C--D--E--F--A--cycle,grey);<br /> fill(arc((2,0),1,0,180)--cycle,white);<br /> fill(arc((2,y),1,180,360)--cycle,white);<br /> fill(arc((z,a),1,60,240)--cycle,white);<br /> fill(arc((b,a),1,300,480)--cycle,white);<br /> fill(arc((b,c),1,240,420)--cycle,white);<br /> fill(arc((z,c),1,120,300)--cycle,white);<br /> draw(A--B--C--D--E--F--A);<br /> draw(arc((z,c),1,120,300));<br /> draw(arc((b,c),1,240,420));<br /> draw(arc((b,a),1,300,480));<br /> draw(arc((z,a),1,60,240));<br /> draw(arc((2,y),1,180,360));<br /> draw(arc((2,0),1,0,180));<br /> label(&quot;2&quot;,(z,c),NE);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi &lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Steve wrote the digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; in order repeatedly from left to right, forming a list of &lt;math&gt;10,000&lt;/math&gt; digits, beginning &lt;math&gt;123451234512\ldots.&lt;/math&gt; He then erased every third digit from his list (that is, the &lt;math&gt;3&lt;/math&gt;rd, &lt;math&gt;6&lt;/math&gt;th, &lt;math&gt;9&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left), then erased every fourth digit from the resulting list (that is, the &lt;math&gt;4&lt;/math&gt;th, &lt;math&gt;8&lt;/math&gt;th, &lt;math&gt;12&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions &lt;math&gt;2019, 2020, 2021&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Bela and Jenn play the following game on the closed interval &lt;math&gt;[0, n]&lt;/math&gt; of the real number line, where &lt;math&gt;n&lt;/math&gt; is a fixed integer greater than &lt;math&gt;4&lt;/math&gt;. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval &lt;math&gt;[0, n]&lt;/math&gt;. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ Bela will always win.} \qquad \textbf{(B)} \text{ Jenn will always win.} \qquad \textbf{(C)} \text{Bela will win if and only if }n \text{ is odd.}&lt;/math&gt;<br /> &lt;math&gt;\textbf{(D)} \text{Jenn will win if and only if }n \text{ is odd.} \qquad \textbf{(E)} \text { Jenn will win if and only if } n&gt;8.&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;B&lt;/math&gt; be a right rectangular prism (box) with edges lengths &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;3,&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, together with its interior. For real &lt;math&gt;r\geq0&lt;/math&gt;, let &lt;math&gt;S(r)&lt;/math&gt; be the set of points in &lt;math&gt;3&lt;/math&gt;-dimensional space that lie within a distance &lt;math&gt;r&lt;/math&gt; of some point &lt;math&gt;B&lt;/math&gt;. The volume of &lt;math&gt;S(r)&lt;/math&gt; can be expressed as &lt;math&gt;ar^{3} + br^{2} + cr +d&lt;/math&gt;, where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive real numbers. What is &lt;math&gt;\frac{bc}{ad}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> In square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively, so that &lt;math&gt;AE=AH.&lt;/math&gt; Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; lie on &lt;math&gt;\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, respectively, and points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lie on &lt;math&gt;\overline{EH}&lt;/math&gt; so that &lt;math&gt;\overline{FI} \perp \overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{GJ} \perp \overline{EH}&lt;/math&gt;. See the figure below. Triangle &lt;math&gt;AEH&lt;/math&gt;, quadrilateral &lt;math&gt;BFIE&lt;/math&gt;, quadrilateral &lt;math&gt;DHJG&lt;/math&gt;, and pentagon &lt;math&gt;FCGJI&lt;/math&gt; each has area &lt;math&gt;1.&lt;/math&gt; What is &lt;math&gt;FI^2&lt;/math&gt;?<br /> &lt;asy&gt;<br /> real x=2sqrt(2);<br /> real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br /> real z=2sqrt(8-4sqrt(2));<br /> pair A, B, C, D, E, F, G, H, I, J;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (4,4);<br /> D = (0,4);<br /> E = (x,0);<br /> F = (4,y);<br /> G = (y,4);<br /> H = (0,x);<br /> I = F + z * dir(225);<br /> J = G + z * dir(225);<br /> <br /> draw(A--B--C--D--A);<br /> draw(H--E);<br /> draw(J--G^^F--I);<br /> draw(rightanglemark(G, J, I), linewidth(.5));<br /> draw(rightanglemark(F, I, E), linewidth(.5));<br /> <br /> dot(&quot;$A$&quot;, A, S);<br /> dot(&quot;$B$&quot;, B, S);<br /> dot(&quot;$C$&quot;, C, dir(90));<br /> dot(&quot;$D$&quot;, D, dir(90));<br /> dot(&quot;$E$&quot;, E, S);<br /> dot(&quot;$F$&quot;, F, dir(0));<br /> dot(&quot;$G$&quot;, G, N);<br /> dot(&quot;$H$&quot;, H, W);<br /> dot(&quot;$I$&quot;, I, SW);<br /> dot(&quot;$J$&quot;, J, SW);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> What is the remainder when &lt;math&gt;2^{202} +202&lt;/math&gt; is divided by &lt;math&gt;2^{101}+2^{51}+1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; in the coordinate plane has vertices at the points &lt;math&gt;A(1,1), B(-1,1), C(-1,-1),&lt;/math&gt; and &lt;math&gt;D(1,-1).&lt;/math&gt; Consider the following four transformations:<br /> &lt;math&gt;L,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise around the origin;<br /> &lt;math&gt;R,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; clockwise around the origin;<br /> &lt;math&gt;H,&lt;/math&gt; a reflection across the &lt;math&gt;x&lt;/math&gt;-axis; and<br /> &lt;math&gt;V,&lt;/math&gt; a reflection across the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying &lt;math&gt;R&lt;/math&gt; and then &lt;math&gt;V&lt;/math&gt; would send the vertex &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;(1,1)&lt;/math&gt; to &lt;math&gt;(-1,-1)&lt;/math&gt; and would send the vertex &lt;math&gt;B&lt;/math&gt; at &lt;math&gt;(-1,1)&lt;/math&gt; to itself. How many sequences of &lt;math&gt;20&lt;/math&gt; transformations chosen from &lt;math&gt;\{L, R, H, V\}&lt;/math&gt; will send all of the labeled vertices back to their original positions? (For example, &lt;math&gt;R, R, V, H&lt;/math&gt; is one sequence of &lt;math&gt;4&lt;/math&gt; transformations that will send the vertices back to their original positions.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy&lt;cmath&gt;\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?&lt;/cmath&gt;(Recall that &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; is the greatest integer not exceeding &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;D(n)&lt;/math&gt; denote the number of ways of writing the positive integer &lt;math&gt;n&lt;/math&gt; as a product&lt;cmath&gt;n = f_1\cdot f_2\cdots f_k,&lt;/cmath&gt;where &lt;math&gt;k\ge1&lt;/math&gt;, the &lt;math&gt;f_i&lt;/math&gt; are integers strictly greater than &lt;math&gt;1&lt;/math&gt;, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number &lt;math&gt;6&lt;/math&gt; can be written as &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2\cdot 3&lt;/math&gt;, and &lt;math&gt;3\cdot2&lt;/math&gt;, so &lt;math&gt;D(6) = 3&lt;/math&gt;. What is &lt;math&gt;D(96)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=B|before=[[2020 AMC 10A Problems]]|after=[[2021 AMC 10A Problems]]}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems&diff=118291 2020 AMC 10B Problems 2020-02-21T23:13:13Z <p>StellarG: /* Problem */</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;1 - (-2) - 3 - (-4) - 5 - (-6)?&lt;/cmath&gt;<br /> <br /> (A) -20 (B) -3 (C) 3 (D) 5 (E) 21<br /> <br /> [[2020 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> Editing 2020 AMC 10B Problems/Problem 2 (section)<br /> <br /> ==Problem 2==<br /> <br /> Aaron has &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;1&lt;/math&gt;, and Steph and Val each have &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;2&lt;/math&gt;. What is the total volume of these &lt;math&gt;10&lt;/math&gt; cubes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The acute angles of a right triangle are &lt;math&gt;a^{\circ}&lt;/math&gt; and &lt;math&gt;b^{\circ}&lt;/math&gt;, where &lt;math&gt;a&gt;b&lt;/math&gt; and both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are prime numbers. What is the least possible value of &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Driving along a highway, Megan noticed that her odometer showed &lt;math&gt;15951&lt;/math&gt; (miles). This number is a palindrome-it reads the same forward and backward. Then &lt;math&gt;2&lt;/math&gt; hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this &lt;math&gt;2&lt;/math&gt;-hour period?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many positive even multiples of &lt;math&gt;3&lt;/math&gt; less than &lt;math&gt;2020&lt;/math&gt; are perfect squares?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie in a plane with &lt;math&gt;PQ=8&lt;/math&gt;. How many locations for point &lt;math&gt;R&lt;/math&gt; in this plane are there such that the triangle with vertices &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; is a right triangle with area &lt;math&gt;12&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> How many ordered pairs of integers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the equation &lt;cmath&gt;x^{2020} + y^2 = 2y?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A three-quarter sector of a circle of radius &lt;math&gt;4&lt;/math&gt; inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?<br /> &lt;asy&gt;<br /> <br /> draw(Arc((0,0), 4, 0, 270));<br /> draw((0,-4)--(0,0)--(4,0));<br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Ms.Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> The decimal representation of &lt;cmath&gt;\frac{1}{20^{20}}&lt;/cmath&gt;<br /> consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Andy the Ant lives on a coordinate plane and is currently at &lt;math&gt;(-20, 20)&lt;/math&gt; facing east (that is, in the positive &lt;math&gt;x&lt;/math&gt;-direction). Andy moves &lt;math&gt;1&lt;/math&gt; unit and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. From there, Andy moves &lt;math&gt;2&lt;/math&gt; units (north) and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. He then moves &lt;math&gt;3&lt;/math&gt; units (west) and again turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. Andy continues his progress, increasing his distance each time by &lt;math&gt;1&lt;/math&gt; unit and always turning left. What is the location of the point at which Andy makes the &lt;math&gt;2020&lt;/math&gt;th left turn?<br /> <br /> &lt;math&gt;\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?<br /> <br /> &lt;asy&gt;<br /> real x=sqrt(3);<br /> real y=2sqrt(3);<br /> real z=3.5;<br /> real a=x/2;<br /> real b=0.5;<br /> real c=3a;<br /> pair A, B, C, D, E, F;<br /> A = (1,0);<br /> B = (3,0);<br /> C = (4,x);<br /> D = (3,y);<br /> E = (1,y);<br /> F = (0,x);<br /> <br /> fill(A--B--C--D--E--F--A--cycle,grey);<br /> fill(arc((2,0),1,0,180)--cycle,white);<br /> fill(arc((2,y),1,180,360)--cycle,white);<br /> fill(arc((z,a),1,60,240)--cycle,white);<br /> fill(arc((b,a),1,300,480)--cycle,white);<br /> fill(arc((b,c),1,240,420)--cycle,white);<br /> fill(arc((z,c),1,120,300)--cycle,white);<br /> draw(A--B--C--D--E--F--A);<br /> draw(arc((z,c),1,120,300));<br /> draw(arc((b,c),1,240,420));<br /> draw(arc((b,a),1,300,480));<br /> draw(arc((z,a),1,60,240));<br /> draw(arc((2,y),1,180,360));<br /> draw(arc((2,0),1,0,180));<br /> label(&quot;2&quot;,(z,c),NE);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi &lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Steve wrote the digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; in order repeatedly from left to right, forming a list of &lt;math&gt;10,000&lt;/math&gt; digits, beginning &lt;math&gt;123451234512\ldots.&lt;/math&gt; He then erased every third digit from his list (that is, the &lt;math&gt;3&lt;/math&gt;rd, &lt;math&gt;6&lt;/math&gt;th, &lt;math&gt;9&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left), then erased every fourth digit from the resulting list (that is, the &lt;math&gt;4&lt;/math&gt;th, &lt;math&gt;8&lt;/math&gt;th, &lt;math&gt;12&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions &lt;math&gt;2019, 2020, 2021&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Bela and Jenn play the following game on the closed interval &lt;math&gt;[0, n]&lt;/math&gt; of the real number line, where &lt;math&gt;n&lt;/math&gt; is a fixed integer greater than &lt;math&gt;4&lt;/math&gt;. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval &lt;math&gt;[0, n]&lt;/math&gt;. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ Bela will always win.} \qquad \textbf{(B)} \text{ Jenn will always win.} \qquad \textbf{(C)} \text{Bela will win if and only if }n \text{ is odd.}&lt;/math&gt;<br /> &lt;math&gt;\textbf{(D)} \text{Jenn will win if and only if }n \text{ is odd.} \qquad \textbf{(E)} \text { Jenn will win if and only if } n&gt;8.&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;B&lt;/math&gt; be a right rectangular prism (box) with edges lengths &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;3,&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, together with its interior. For real &lt;math&gt;r\geq0&lt;/math&gt;, let &lt;math&gt;S(r)&lt;/math&gt; be the set of points in &lt;math&gt;3&lt;/math&gt;-dimensional space that lie within a distance &lt;math&gt;r&lt;/math&gt; of some point &lt;math&gt;B&lt;/math&gt;. The volume of &lt;math&gt;S(r)&lt;/math&gt; can be expressed as &lt;math&gt;ar^{3} + br^{2} + cr +d&lt;/math&gt;, where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive real numbers. What is &lt;math&gt;\frac{bc}{ad}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> In square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively, so that &lt;math&gt;AE=AH.&lt;/math&gt; Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; lie on &lt;math&gt;\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, respectively, and points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lie on &lt;math&gt;\overline{EH}&lt;/math&gt; so that &lt;math&gt;\overline{FI} \perp \overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{GJ} \perp \overline{EH}&lt;/math&gt;. See the figure below. Triangle &lt;math&gt;AEH&lt;/math&gt;, quadrilateral &lt;math&gt;BFIE&lt;/math&gt;, quadrilateral &lt;math&gt;DHJG&lt;/math&gt;, and pentagon &lt;math&gt;FCGJI&lt;/math&gt; each has area &lt;math&gt;1.&lt;/math&gt; What is &lt;math&gt;FI^2&lt;/math&gt;?<br /> &lt;asy&gt;<br /> real x=2sqrt(2);<br /> real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br /> real z=2sqrt(8-4sqrt(2));<br /> pair A, B, C, D, E, F, G, H, I, J;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (4,4);<br /> D = (0,4);<br /> E = (x,0);<br /> F = (4,y);<br /> G = (y,4);<br /> H = (0,x);<br /> I = F + z * dir(225);<br /> J = G + z * dir(225);<br /> <br /> draw(A--B--C--D--A);<br /> draw(H--E);<br /> draw(J--G^^F--I);<br /> draw(rightanglemark(G, J, I), linewidth(.5));<br /> draw(rightanglemark(F, I, E), linewidth(.5));<br /> <br /> dot(&quot;$A$&quot;, A, S);<br /> dot(&quot;$B$&quot;, B, S);<br /> dot(&quot;$C$&quot;, C, dir(90));<br /> dot(&quot;$D$&quot;, D, dir(90));<br /> dot(&quot;$E$&quot;, E, S);<br /> dot(&quot;$F$&quot;, F, dir(0));<br /> dot(&quot;$G$&quot;, G, N);<br /> dot(&quot;$H$&quot;, H, W);<br /> dot(&quot;$I$&quot;, I, SW);<br /> dot(&quot;$J$&quot;, J, SW);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> What is the remainder when &lt;math&gt;2^{202} +202&lt;/math&gt; is divided by &lt;math&gt;2^{101}+2^{51}+1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; in the coordinate plane has vertices at the points &lt;math&gt;A(1,1), B(-1,1), C(-1,-1),&lt;/math&gt; and &lt;math&gt;D(1,-1).&lt;/math&gt; Consider the following four transformations:<br /> &lt;math&gt;L,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise around the origin;<br /> &lt;math&gt;R,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; clockwise around the origin;<br /> &lt;math&gt;H,&lt;/math&gt; a reflection across the &lt;math&gt;x&lt;/math&gt;-axis; and<br /> &lt;math&gt;V,&lt;/math&gt; a reflection across the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying &lt;math&gt;R&lt;/math&gt; and then &lt;math&gt;V&lt;/math&gt; would send the vertex &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;(1,1)&lt;/math&gt; to &lt;math&gt;(-1,-1)&lt;/math&gt; and would send the vertex &lt;math&gt;B&lt;/math&gt; at &lt;math&gt;(-1,1)&lt;/math&gt; to itself. How many sequences of &lt;math&gt;20&lt;/math&gt; transformations chosen from &lt;math&gt;\{L, R, H, V\}&lt;/math&gt; will send all of the labeled vertices back to their original positions? (For example, &lt;math&gt;R, R, V, H&lt;/math&gt; is one sequence of &lt;math&gt;4&lt;/math&gt; transformations that will send the vertices back to their original positions.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy&lt;cmath&gt;\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?&lt;/cmath&gt;(Recall that &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; is the greatest integer not exceeding &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;D(n)&lt;/math&gt; denote the number of ways of writing the positive integer &lt;math&gt;n&lt;/math&gt; as a product&lt;cmath&gt;n = f_1\cdot f_2\cdots f_k,&lt;/cmath&gt;where &lt;math&gt;k\ge1&lt;/math&gt;, the &lt;math&gt;f_i&lt;/math&gt; are integers strictly greater than &lt;math&gt;1&lt;/math&gt;, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number &lt;math&gt;6&lt;/math&gt; can be written as &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2\cdot 3&lt;/math&gt;, and &lt;math&gt;3\cdot2&lt;/math&gt;, so &lt;math&gt;D(6) = 3&lt;/math&gt;. What is &lt;math&gt;D(96)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=B|before=[[2020 AMC 10A Problems]]|after=[[2021 AMC 10A Problems]]}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems&diff=118290 2020 AMC 10B Problems 2020-02-21T23:12:08Z <p>StellarG: /* Problem 1 */</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;1 - (-2) - 3 - (-4) - 5 - (-6)?&lt;/cmath&gt;<br /> <br /> (A) -20 (B) -3 (C) 3 (D) 5 (E) 21<br /> <br /> [[2020 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> Editing 2020 AMC 10B Problems/Problem 2 (section)<br /> <br /> ==Problem==<br /> <br /> Aaron has &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;1&lt;/math&gt;, and Steph and Val each have &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;2&lt;/math&gt;. What is the total volume of these &lt;math&gt;10&lt;/math&gt; cubes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7777777 \qquad\textbf{(B)}\ 4444 \qquad\textbf{(C)}\ 333 \qquad\textbf{(D)}\ None \qquad\textbf{(E)}\ NONE&lt;/math&gt;<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The acute angles of a right triangle are &lt;math&gt;a^{\circ}&lt;/math&gt; and &lt;math&gt;b^{\circ}&lt;/math&gt;, where &lt;math&gt;a&gt;b&lt;/math&gt; and both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are prime numbers. What is the least possible value of &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Driving along a highway, Megan noticed that her odometer showed &lt;math&gt;15951&lt;/math&gt; (miles). This number is a palindrome-it reads the same forward and backward. Then &lt;math&gt;2&lt;/math&gt; hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this &lt;math&gt;2&lt;/math&gt;-hour period?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many positive even multiples of &lt;math&gt;3&lt;/math&gt; less than &lt;math&gt;2020&lt;/math&gt; are perfect squares?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie in a plane with &lt;math&gt;PQ=8&lt;/math&gt;. How many locations for point &lt;math&gt;R&lt;/math&gt; in this plane are there such that the triangle with vertices &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; is a right triangle with area &lt;math&gt;12&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> How many ordered pairs of integers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the equation &lt;cmath&gt;x^{2020} + y^2 = 2y?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A three-quarter sector of a circle of radius &lt;math&gt;4&lt;/math&gt; inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?<br /> &lt;asy&gt;<br /> <br /> draw(Arc((0,0), 4, 0, 270));<br /> draw((0,-4)--(0,0)--(4,0));<br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Ms.Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> The decimal representation of &lt;cmath&gt;\frac{1}{20^{20}}&lt;/cmath&gt;<br /> consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Andy the Ant lives on a coordinate plane and is currently at &lt;math&gt;(-20, 20)&lt;/math&gt; facing east (that is, in the positive &lt;math&gt;x&lt;/math&gt;-direction). Andy moves &lt;math&gt;1&lt;/math&gt; unit and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. From there, Andy moves &lt;math&gt;2&lt;/math&gt; units (north) and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. He then moves &lt;math&gt;3&lt;/math&gt; units (west) and again turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. Andy continues his progress, increasing his distance each time by &lt;math&gt;1&lt;/math&gt; unit and always turning left. What is the location of the point at which Andy makes the &lt;math&gt;2020&lt;/math&gt;th left turn?<br /> <br /> &lt;math&gt;\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?<br /> <br /> &lt;asy&gt;<br /> real x=sqrt(3);<br /> real y=2sqrt(3);<br /> real z=3.5;<br /> real a=x/2;<br /> real b=0.5;<br /> real c=3a;<br /> pair A, B, C, D, E, F;<br /> A = (1,0);<br /> B = (3,0);<br /> C = (4,x);<br /> D = (3,y);<br /> E = (1,y);<br /> F = (0,x);<br /> <br /> fill(A--B--C--D--E--F--A--cycle,grey);<br /> fill(arc((2,0),1,0,180)--cycle,white);<br /> fill(arc((2,y),1,180,360)--cycle,white);<br /> fill(arc((z,a),1,60,240)--cycle,white);<br /> fill(arc((b,a),1,300,480)--cycle,white);<br /> fill(arc((b,c),1,240,420)--cycle,white);<br /> fill(arc((z,c),1,120,300)--cycle,white);<br /> draw(A--B--C--D--E--F--A);<br /> draw(arc((z,c),1,120,300));<br /> draw(arc((b,c),1,240,420));<br /> draw(arc((b,a),1,300,480));<br /> draw(arc((z,a),1,60,240));<br /> draw(arc((2,y),1,180,360));<br /> draw(arc((2,0),1,0,180));<br /> label(&quot;2&quot;,(z,c),NE);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi &lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Steve wrote the digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; in order repeatedly from left to right, forming a list of &lt;math&gt;10,000&lt;/math&gt; digits, beginning &lt;math&gt;123451234512\ldots.&lt;/math&gt; He then erased every third digit from his list (that is, the &lt;math&gt;3&lt;/math&gt;rd, &lt;math&gt;6&lt;/math&gt;th, &lt;math&gt;9&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left), then erased every fourth digit from the resulting list (that is, the &lt;math&gt;4&lt;/math&gt;th, &lt;math&gt;8&lt;/math&gt;th, &lt;math&gt;12&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions &lt;math&gt;2019, 2020, 2021&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Bela and Jenn play the following game on the closed interval &lt;math&gt;[0, n]&lt;/math&gt; of the real number line, where &lt;math&gt;n&lt;/math&gt; is a fixed integer greater than &lt;math&gt;4&lt;/math&gt;. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval &lt;math&gt;[0, n]&lt;/math&gt;. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ Bela will always win.} \qquad \textbf{(B)} \text{ Jenn will always win.} \qquad \textbf{(C)} \text{Bela will win if and only if }n \text{ is odd.}&lt;/math&gt;<br /> &lt;math&gt;\textbf{(D)} \text{Jenn will win if and only if }n \text{ is odd.} \qquad \textbf{(E)} \text { Jenn will win if and only if } n&gt;8.&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;B&lt;/math&gt; be a right rectangular prism (box) with edges lengths &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;3,&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, together with its interior. For real &lt;math&gt;r\geq0&lt;/math&gt;, let &lt;math&gt;S(r)&lt;/math&gt; be the set of points in &lt;math&gt;3&lt;/math&gt;-dimensional space that lie within a distance &lt;math&gt;r&lt;/math&gt; of some point &lt;math&gt;B&lt;/math&gt;. The volume of &lt;math&gt;S(r)&lt;/math&gt; can be expressed as &lt;math&gt;ar^{3} + br^{2} + cr +d&lt;/math&gt;, where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive real numbers. What is &lt;math&gt;\frac{bc}{ad}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> In square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively, so that &lt;math&gt;AE=AH.&lt;/math&gt; Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; lie on &lt;math&gt;\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, respectively, and points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lie on &lt;math&gt;\overline{EH}&lt;/math&gt; so that &lt;math&gt;\overline{FI} \perp \overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{GJ} \perp \overline{EH}&lt;/math&gt;. See the figure below. Triangle &lt;math&gt;AEH&lt;/math&gt;, quadrilateral &lt;math&gt;BFIE&lt;/math&gt;, quadrilateral &lt;math&gt;DHJG&lt;/math&gt;, and pentagon &lt;math&gt;FCGJI&lt;/math&gt; each has area &lt;math&gt;1.&lt;/math&gt; What is &lt;math&gt;FI^2&lt;/math&gt;?<br /> &lt;asy&gt;<br /> real x=2sqrt(2);<br /> real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br /> real z=2sqrt(8-4sqrt(2));<br /> pair A, B, C, D, E, F, G, H, I, J;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (4,4);<br /> D = (0,4);<br /> E = (x,0);<br /> F = (4,y);<br /> G = (y,4);<br /> H = (0,x);<br /> I = F + z * dir(225);<br /> J = G + z * dir(225);<br /> <br /> draw(A--B--C--D--A);<br /> draw(H--E);<br /> draw(J--G^^F--I);<br /> draw(rightanglemark(G, J, I), linewidth(.5));<br /> draw(rightanglemark(F, I, E), linewidth(.5));<br /> <br /> dot(&quot;$A$&quot;, A, S);<br /> dot(&quot;$B$&quot;, B, S);<br /> dot(&quot;$C$&quot;, C, dir(90));<br /> dot(&quot;$D$&quot;, D, dir(90));<br /> dot(&quot;$E$&quot;, E, S);<br /> dot(&quot;$F$&quot;, F, dir(0));<br /> dot(&quot;$G$&quot;, G, N);<br /> dot(&quot;$H$&quot;, H, W);<br /> dot(&quot;$I$&quot;, I, SW);<br /> dot(&quot;$J$&quot;, J, SW);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> What is the remainder when &lt;math&gt;2^{202} +202&lt;/math&gt; is divided by &lt;math&gt;2^{101}+2^{51}+1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; in the coordinate plane has vertices at the points &lt;math&gt;A(1,1), B(-1,1), C(-1,-1),&lt;/math&gt; and &lt;math&gt;D(1,-1).&lt;/math&gt; Consider the following four transformations:<br /> &lt;math&gt;L,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise around the origin;<br /> &lt;math&gt;R,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; clockwise around the origin;<br /> &lt;math&gt;H,&lt;/math&gt; a reflection across the &lt;math&gt;x&lt;/math&gt;-axis; and<br /> &lt;math&gt;V,&lt;/math&gt; a reflection across the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying &lt;math&gt;R&lt;/math&gt; and then &lt;math&gt;V&lt;/math&gt; would send the vertex &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;(1,1)&lt;/math&gt; to &lt;math&gt;(-1,-1)&lt;/math&gt; and would send the vertex &lt;math&gt;B&lt;/math&gt; at &lt;math&gt;(-1,1)&lt;/math&gt; to itself. How many sequences of &lt;math&gt;20&lt;/math&gt; transformations chosen from &lt;math&gt;\{L, R, H, V\}&lt;/math&gt; will send all of the labeled vertices back to their original positions? (For example, &lt;math&gt;R, R, V, H&lt;/math&gt; is one sequence of &lt;math&gt;4&lt;/math&gt; transformations that will send the vertices back to their original positions.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy&lt;cmath&gt;\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?&lt;/cmath&gt;(Recall that &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; is the greatest integer not exceeding &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;D(n)&lt;/math&gt; denote the number of ways of writing the positive integer &lt;math&gt;n&lt;/math&gt; as a product&lt;cmath&gt;n = f_1\cdot f_2\cdots f_k,&lt;/cmath&gt;where &lt;math&gt;k\ge1&lt;/math&gt;, the &lt;math&gt;f_i&lt;/math&gt; are integers strictly greater than &lt;math&gt;1&lt;/math&gt;, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number &lt;math&gt;6&lt;/math&gt; can be written as &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2\cdot 3&lt;/math&gt;, and &lt;math&gt;3\cdot2&lt;/math&gt;, so &lt;math&gt;D(6) = 3&lt;/math&gt;. What is &lt;math&gt;D(96)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=B|before=[[2020 AMC 10A Problems]]|after=[[2021 AMC 10A Problems]]}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems&diff=118289 2020 AMC 10B Problems 2020-02-21T22:58:17Z <p>StellarG: /* Problem 1 */</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=B}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;cmath&gt;1 - (-2) - 3 - (-4) - 5 - (-6)?&lt;/cmath&gt;<br /> <br /> <br /> <br /> [[2020 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> Editing 2020 AMC 10B Problems/Problem 2 (section)<br /> <br /> ==Problem==<br /> <br /> Aaron has &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;1&lt;/math&gt;, and Steph and Val each have &lt;math&gt;5&lt;/math&gt; cubes each having side length &lt;math&gt;2&lt;/math&gt;. What is the total volume of these &lt;math&gt;10&lt;/math&gt; cubes?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7777777 \qquad\textbf{(B)}\ 4444 \qquad\textbf{(C)}\ 333 \qquad\textbf{(D)}\ None \qquad\textbf{(E)}\ NONE&lt;/math&gt;<br /> <br /> ==Problem 3==<br /> <br /> The ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;4:3&lt;/math&gt;, the ratio of &lt;math&gt;y&lt;/math&gt; to &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;3:2&lt;/math&gt;, and the ratio of &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;1:6&lt;/math&gt;. What is the ratio of &lt;math&gt;w&lt;/math&gt; to &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The acute angles of a right triangle are &lt;math&gt;a^{\circ}&lt;/math&gt; and &lt;math&gt;b^{\circ}&lt;/math&gt;, where &lt;math&gt;a&gt;b&lt;/math&gt; and both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are prime numbers. What is the least possible value of &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Driving along a highway, Megan noticed that her odometer showed &lt;math&gt;15951&lt;/math&gt; (miles). This number is a palindrome-it reads the same forward and backward. Then &lt;math&gt;2&lt;/math&gt; hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this &lt;math&gt;2&lt;/math&gt;-hour period?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> How many positive even multiples of &lt;math&gt;3&lt;/math&gt; less than &lt;math&gt;2020&lt;/math&gt; are perfect squares?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; lie in a plane with &lt;math&gt;PQ=8&lt;/math&gt;. How many locations for point &lt;math&gt;R&lt;/math&gt; in this plane are there such that the triangle with vertices &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, and &lt;math&gt;R&lt;/math&gt; is a right triangle with area &lt;math&gt;12&lt;/math&gt; square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> How many ordered pairs of integers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the equation &lt;cmath&gt;x^{2020} + y^2 = 2y?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A three-quarter sector of a circle of radius &lt;math&gt;4&lt;/math&gt; inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?<br /> &lt;asy&gt;<br /> <br /> draw(Arc((0,0), 4, 0, 270));<br /> draw((0,-4)--(0,0)--(4,0));<br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Ms.Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> The decimal representation of &lt;cmath&gt;\frac{1}{20^{20}}&lt;/cmath&gt;<br /> consists of a string of zeros after the decimal point, followed by a &lt;math&gt;9&lt;/math&gt; and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Andy the Ant lives on a coordinate plane and is currently at &lt;math&gt;(-20, 20)&lt;/math&gt; facing east (that is, in the positive &lt;math&gt;x&lt;/math&gt;-direction). Andy moves &lt;math&gt;1&lt;/math&gt; unit and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. From there, Andy moves &lt;math&gt;2&lt;/math&gt; units (north) and then turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. He then moves &lt;math&gt;3&lt;/math&gt; units (west) and again turns &lt;math&gt;90^{\circ}&lt;/math&gt; degrees left. Andy continues his progress, increasing his distance each time by &lt;math&gt;1&lt;/math&gt; unit and always turning left. What is the location of the point at which Andy makes the &lt;math&gt;2020&lt;/math&gt;th left turn?<br /> <br /> &lt;math&gt;\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?<br /> <br /> &lt;asy&gt;<br /> real x=sqrt(3);<br /> real y=2sqrt(3);<br /> real z=3.5;<br /> real a=x/2;<br /> real b=0.5;<br /> real c=3a;<br /> pair A, B, C, D, E, F;<br /> A = (1,0);<br /> B = (3,0);<br /> C = (4,x);<br /> D = (3,y);<br /> E = (1,y);<br /> F = (0,x);<br /> <br /> fill(A--B--C--D--E--F--A--cycle,grey);<br /> fill(arc((2,0),1,0,180)--cycle,white);<br /> fill(arc((2,y),1,180,360)--cycle,white);<br /> fill(arc((z,a),1,60,240)--cycle,white);<br /> fill(arc((b,a),1,300,480)--cycle,white);<br /> fill(arc((b,c),1,240,420)--cycle,white);<br /> fill(arc((z,c),1,120,300)--cycle,white);<br /> draw(A--B--C--D--E--F--A);<br /> draw(arc((z,c),1,120,300));<br /> draw(arc((b,c),1,240,420));<br /> draw(arc((b,a),1,300,480));<br /> draw(arc((z,a),1,60,240));<br /> draw(arc((2,y),1,180,360));<br /> draw(arc((2,0),1,0,180));<br /> label(&quot;2&quot;,(z,c),NE);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi &lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Steve wrote the digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; in order repeatedly from left to right, forming a list of &lt;math&gt;10,000&lt;/math&gt; digits, beginning &lt;math&gt;123451234512\ldots.&lt;/math&gt; He then erased every third digit from his list (that is, the &lt;math&gt;3&lt;/math&gt;rd, &lt;math&gt;6&lt;/math&gt;th, &lt;math&gt;9&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left), then erased every fourth digit from the resulting list (that is, the &lt;math&gt;4&lt;/math&gt;th, &lt;math&gt;8&lt;/math&gt;th, &lt;math&gt;12&lt;/math&gt;th, &lt;math&gt;\ldots&lt;/math&gt; digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions &lt;math&gt;2019, 2020, 2021&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Bela and Jenn play the following game on the closed interval &lt;math&gt;[0, n]&lt;/math&gt; of the real number line, where &lt;math&gt;n&lt;/math&gt; is a fixed integer greater than &lt;math&gt;4&lt;/math&gt;. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval &lt;math&gt;[0, n]&lt;/math&gt;. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ Bela will always win.} \qquad \textbf{(B)} \text{ Jenn will always win.} \qquad \textbf{(C)} \text{Bela will win if and only if }n \text{ is odd.}&lt;/math&gt;<br /> &lt;math&gt;\textbf{(D)} \text{Jenn will win if and only if }n \text{ is odd.} \qquad \textbf{(E)} \text { Jenn will win if and only if } n&gt;8.&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;B&lt;/math&gt; be a right rectangular prism (box) with edges lengths &lt;math&gt;1,&lt;/math&gt; &lt;math&gt;3,&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, together with its interior. For real &lt;math&gt;r\geq0&lt;/math&gt;, let &lt;math&gt;S(r)&lt;/math&gt; be the set of points in &lt;math&gt;3&lt;/math&gt;-dimensional space that lie within a distance &lt;math&gt;r&lt;/math&gt; of some point &lt;math&gt;B&lt;/math&gt;. The volume of &lt;math&gt;S(r)&lt;/math&gt; can be expressed as &lt;math&gt;ar^{3} + br^{2} + cr +d&lt;/math&gt;, where &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;b,&lt;/math&gt; &lt;math&gt;c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are positive real numbers. What is &lt;math&gt;\frac{bc}{ad}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> In square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively, so that &lt;math&gt;AE=AH.&lt;/math&gt; Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; lie on &lt;math&gt;\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, respectively, and points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lie on &lt;math&gt;\overline{EH}&lt;/math&gt; so that &lt;math&gt;\overline{FI} \perp \overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{GJ} \perp \overline{EH}&lt;/math&gt;. See the figure below. Triangle &lt;math&gt;AEH&lt;/math&gt;, quadrilateral &lt;math&gt;BFIE&lt;/math&gt;, quadrilateral &lt;math&gt;DHJG&lt;/math&gt;, and pentagon &lt;math&gt;FCGJI&lt;/math&gt; each has area &lt;math&gt;1.&lt;/math&gt; What is &lt;math&gt;FI^2&lt;/math&gt;?<br /> &lt;asy&gt;<br /> real x=2sqrt(2);<br /> real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br /> real z=2sqrt(8-4sqrt(2));<br /> pair A, B, C, D, E, F, G, H, I, J;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (4,4);<br /> D = (0,4);<br /> E = (x,0);<br /> F = (4,y);<br /> G = (y,4);<br /> H = (0,x);<br /> I = F + z * dir(225);<br /> J = G + z * dir(225);<br /> <br /> draw(A--B--C--D--A);<br /> draw(H--E);<br /> draw(J--G^^F--I);<br /> draw(rightanglemark(G, J, I), linewidth(.5));<br /> draw(rightanglemark(F, I, E), linewidth(.5));<br /> <br /> dot(&quot;$A$&quot;, A, S);<br /> dot(&quot;$B$&quot;, B, S);<br /> dot(&quot;$C$&quot;, C, dir(90));<br /> dot(&quot;$D$&quot;, D, dir(90));<br /> dot(&quot;$E$&quot;, E, S);<br /> dot(&quot;$F$&quot;, F, dir(0));<br /> dot(&quot;$G$&quot;, G, N);<br /> dot(&quot;$H$&quot;, H, W);<br /> dot(&quot;$I$&quot;, I, SW);<br /> dot(&quot;$J$&quot;, J, SW);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> What is the remainder when &lt;math&gt;2^{202} +202&lt;/math&gt; is divided by &lt;math&gt;2^{101}+2^{51}+1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Square &lt;math&gt;ABCD&lt;/math&gt; in the coordinate plane has vertices at the points &lt;math&gt;A(1,1), B(-1,1), C(-1,-1),&lt;/math&gt; and &lt;math&gt;D(1,-1).&lt;/math&gt; Consider the following four transformations:<br /> &lt;math&gt;L,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; counterclockwise around the origin;<br /> &lt;math&gt;R,&lt;/math&gt; a rotation of &lt;math&gt;90^{\circ}&lt;/math&gt; clockwise around the origin;<br /> &lt;math&gt;H,&lt;/math&gt; a reflection across the &lt;math&gt;x&lt;/math&gt;-axis; and<br /> &lt;math&gt;V,&lt;/math&gt; a reflection across the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying &lt;math&gt;R&lt;/math&gt; and then &lt;math&gt;V&lt;/math&gt; would send the vertex &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;(1,1)&lt;/math&gt; to &lt;math&gt;(-1,-1)&lt;/math&gt; and would send the vertex &lt;math&gt;B&lt;/math&gt; at &lt;math&gt;(-1,1)&lt;/math&gt; to itself. How many sequences of &lt;math&gt;20&lt;/math&gt; transformations chosen from &lt;math&gt;\{L, R, H, V\}&lt;/math&gt; will send all of the labeled vertices back to their original positions? (For example, &lt;math&gt;R, R, V, H&lt;/math&gt; is one sequence of &lt;math&gt;4&lt;/math&gt; transformations that will send the vertices back to their original positions.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> How many positive integers &lt;math&gt;n&lt;/math&gt; satisfy&lt;cmath&gt;\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?&lt;/cmath&gt;(Recall that &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; is the greatest integer not exceeding &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;D(n)&lt;/math&gt; denote the number of ways of writing the positive integer &lt;math&gt;n&lt;/math&gt; as a product&lt;cmath&gt;n = f_1\cdot f_2\cdots f_k,&lt;/cmath&gt;where &lt;math&gt;k\ge1&lt;/math&gt;, the &lt;math&gt;f_i&lt;/math&gt; are integers strictly greater than &lt;math&gt;1&lt;/math&gt;, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number &lt;math&gt;6&lt;/math&gt; can be written as &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2\cdot 3&lt;/math&gt;, and &lt;math&gt;3\cdot2&lt;/math&gt;, so &lt;math&gt;D(6) = 3&lt;/math&gt;. What is &lt;math&gt;D(96)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184&lt;/math&gt;<br /> <br /> [[2020 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=B|before=[[2020 AMC 10A Problems]]|after=[[2021 AMC 10A Problems]]}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_3&diff=98781 2018 AMC 8 Problems/Problem 3 2018-11-21T18:32:23Z <p>StellarG: /* Problem 3 */</p> <hr /> <div>==Problem 3==<br /> Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The five numbers which cause people to leave the circle are &lt;math&gt;7, 14, 17, 21,&lt;/math&gt; and &lt;math&gt;27.&lt;/math&gt;<br /> <br /> Arn counts &lt;math&gt;7&lt;/math&gt; (assuming they start at &lt;math&gt;1&lt;/math&gt;) so he leaves first. Then Cyd counts &lt;math&gt;14&lt;/math&gt;, as there are &lt;math&gt;7&lt;/math&gt; numbers to be counted from this point. Then Fon, Bob, Eve, so last one standing is Dan, &lt;math&gt;\textbf{(D)}&lt;/math&gt; -scrabbler94<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_3&diff=98780 2018 AMC 8 Problems/Problem 3 2018-11-21T18:32:01Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 3==<br /> Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?<br /> <br /> <br /> ==Solution==<br /> The five numbers which cause people to leave the circle are &lt;math&gt;7, 14, 17, 21,&lt;/math&gt; and &lt;math&gt;27.&lt;/math&gt;<br /> <br /> Arn counts &lt;math&gt;7&lt;/math&gt; (assuming they start at &lt;math&gt;1&lt;/math&gt;) so he leaves first. Then Cyd counts &lt;math&gt;14&lt;/math&gt;, as there are &lt;math&gt;7&lt;/math&gt; numbers to be counted from this point. Then Fon, Bob, Eve, so last one standing is Dan, &lt;math&gt;\textbf{(D)}&lt;/math&gt; -scrabbler94<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_13&diff=98109 2000 AMC 10 Problems/Problem 13 2018-10-12T23:35:19Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?<br /> <br /> &lt;asy&gt;<br /> unitsize(20);<br /> dot((0,0));<br /> dot((1,0));<br /> dot((2,0));<br /> dot((3,0));<br /> dot((4,0));<br /> dot((0,1));<br /> dot((1,1));<br /> dot((2,1));<br /> dot((3,1));<br /> dot((0,2));<br /> dot((1,2));<br /> dot((2,2));<br /> dot((0,3));<br /> dot((1,3));<br /> dot((0,4));<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 5!\cdot 4!\cdot 3!\cdot 2!\cdot 1! \qquad\mathrm{(D)}\ \frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!} \qquad\mathrm{(E)}\ 15!&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there. <br /> <br /> In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.<br /> <br /> By similar logic, we can fill in the yellow pegs as shown:<br /> <br /> &lt;asy&gt;<br /> unitsize(20);<br /> dot((0,0));<br /> dot((1,0));<br /> dot((2,0));<br /> dot((3,0));<br /> label(&quot;Y&quot;,(4,-.35),N);<br /> dot((0,1));<br /> dot((1,1));<br /> dot((2,1));<br /> label(&quot;Y&quot;,(3,.6),N);<br /> dot((0,2));<br /> dot((1,2));<br /> label(&quot;Y&quot;,(2,1.6),N);<br /> dot((0,3));<br /> label(&quot;Y&quot;,(1,2.6),N);<br /> label(&quot;Y&quot;,(0,3.6),N);<br /> &lt;/asy&gt;<br /> <br /> After this we can proceed to fill in the whole pegboard, so there is only &lt;math&gt;1&lt;/math&gt; arrangement of the pegs. The answer is &lt;math&gt;\boxed{\text{B}}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(20);<br /> label(&quot;O&quot;,(0,-.35),N);<br /> label(&quot;B&quot;,(1,-.35),N);<br /> label(&quot;G&quot;,(2,-.35),N);<br /> label(&quot;R&quot;,(3,-.35),N);<br /> label(&quot;Y&quot;,(4,-.35),N);<br /> label(&quot;B&quot;,(0,.6),N);<br /> label(&quot;G&quot;,(1,.6),N);<br /> label(&quot;R&quot;,(2,.6),N);<br /> label(&quot;Y&quot;,(3,.6),N);<br /> label(&quot;G&quot;,(0,1.6),N);<br /> label(&quot;R&quot;,(1,1.6),N);<br /> label(&quot;Y&quot;,(2,1.6),N);<br /> label(&quot;R&quot;,(0,2.6),N);<br /> label(&quot;Y&quot;,(1,2.6),N);<br /> label(&quot;Y&quot;,(0,3.6),N);<br /> &lt;/asy&gt;<br /> If (but this is not the case) we were told to distinguish pegs of the same color, the answer would be C.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2000|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_21&diff=91107 2018 AMC 10A Problems/Problem 21 2018-02-12T14:36:14Z <p>StellarG: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> Which of the following describes the set of values of &lt;math&gt;a&lt;/math&gt; for which the curves &lt;math&gt;x^2+y^2=a^2&lt;/math&gt; and &lt;math&gt;y=x^2-a&lt;/math&gt; in the real &lt;math&gt;xy&lt;/math&gt;-plane intersect at exactly &lt;math&gt;3&lt;/math&gt; points?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }a=\frac14 \qquad<br /> \textbf{(B) }\frac14 &lt; a &lt; \frac12 \qquad<br /> \textbf{(C) }a&gt;\frac14 \qquad<br /> \textbf{(D) }a=\frac12 \qquad<br /> \textbf{(E) }a&gt;\frac12 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Substituting &lt;math&gt;y=x^2-a&lt;/math&gt; into &lt;math&gt;x^2+y^2=a^2&lt;/math&gt;, we get<br /> &lt;cmath&gt;<br /> x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0<br /> &lt;/cmath&gt;<br /> Since this is a quartic, there are 4 total roots (counting multiplicity). We see that &lt;math&gt;x=0&lt;/math&gt; always at least one intersection at &lt;math&gt;(0,-a)&lt;/math&gt; (and is in fact a double root). <br /> <br /> The other two intersection points have &lt;math&gt;x&lt;/math&gt; coordinates &lt;math&gt;\sqrt{2a-1}&lt;/math&gt;. We must have &lt;math&gt;2a-1&gt; 0,&lt;/math&gt; otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point &lt;math&gt;(0,a)&lt;/math&gt;). This only results in a single intersection point in the real coordinate plane. Thus, we see &lt;math&gt;a&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> (projecteulerlover)<br /> <br /> == Solution 2 ==<br /> <br /> &lt;asy&gt;<br /> Label f; <br /> f.p=fontsize(6);<br /> xaxis(-2,2,Ticks(f, 0.2)); <br /> yaxis(-2,2,Ticks(f, 0.2)); <br /> real g(real x) <br /> { <br /> return x^2-1; <br /> } <br /> draw(graph(g, 1.7, -1.7));<br /> real h(real x) <br /> { <br /> return sqrt(1-x^2); <br /> } <br /> draw(graph(h, 1, -1));<br /> real j(real x) <br /> { <br /> return -sqrt(1-x^2); <br /> } <br /> draw(graph(j, 1, -1));<br /> &lt;/asy&gt;<br /> <br /> Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola go's 'in' the circle, than by going out of it (as it will) it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have &lt;math&gt;x^2 - a = -\sqrt(a^2 - x^2)&lt;/math&gt;. Squaring both sides and solving yields &lt;math&gt;x^4 - (2a - 1)x^2 = 0&lt;/math&gt;. Since x = 0 is already accounted for, we only need to find 1 solution for &lt;math&gt;x^2 = 2a - 1&lt;/math&gt;, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have &lt;math&gt;a &gt; 1/2 -&gt; E&lt;/math&gt; is the right answer.<br /> <br /> Solution by JohnHankock<br /> <br /> == Solution 3 ==<br /> <br /> This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, &lt;math&gt;\boxed{a &gt; \frac{1}{2}}&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt; is correct.<br /> <br /> &lt;math&gt;QED \blacksquare&lt;/math&gt;<br /> <br /> == Solution 4 ==<br /> <br /> Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point &lt;math&gt;(0, a)&lt;/math&gt;, and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution. <br /> <br /> First, rewrite the second equation to &lt;math&gt;y=x^2-a\implies x^2=y+a&lt;/math&gt;<br /> And substitute into the first equation: &lt;math&gt;y+a+y^2=a^2&lt;/math&gt; <br /> Since we're only interested in seeing the interval in which a can exist, we find the discriminant: &lt;math&gt;1-4a+4a^2&lt;/math&gt;. This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: <br /> &lt;cmath&gt;4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}&lt;/cmath&gt;<br /> Since &lt;math&gt;\lim_{a\to \infty}(4a^2-4a+1)=\infty&lt;/math&gt;, our range is &lt;math&gt;\boxed{a&gt;\frac{1}{2}}&lt;/math&gt;<br /> <br /> Solution by ktong<br /> <br /> == Solution 5 (Cheating with Answer Choices) ==<br /> Simply plug in &lt;math&gt;a = 0, \frac{1}{2}, \frac{1}{4}, 1&lt;/math&gt; and solve the systems. (This shouldn't take too long.) And realized that only &lt;math&gt;a=1&lt;/math&gt; yields three real solutions for &lt;math&gt;x&lt;/math&gt;, so we are done and the answer is &lt;math&gt;\boxed{a&gt;\frac{1}{2}}&lt;/math&gt;<br /> <br /> ~ ccx09<br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=88472 2017 AMC 8 Problems/Problem 9 2017-11-22T19:27:03Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 9==<br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The 6 green marbles are part of &lt;math&gt;1 - \frac13 - \frac14 = \frac5{12}&lt;/math&gt; of the total marbles. If &lt;math&gt;6 \implies \frac13&lt;/math&gt; of the total number of marbles, then there would be 18 marbles. Since a fourth of 18 is not a whole number, we cannot have 18 marbles. Then in &lt;math&gt;6 \implies \frac14&lt;/math&gt; of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 6 - 8 = 10 - 8 = 4 marbles, or D.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_18&diff=88471 2017 AMC 8 Problems/Problem 18 2017-11-22T19:26:44Z <p>StellarG: Created page with &quot;==Problem 18== In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math...&quot;</p> <hr /> <div>==Problem 18==<br /> In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math&gt;, and &lt;math&gt;AD=13&lt;/math&gt;. [asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;, (0, 0), SW); label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;, (12, 0), ESE); label(&quot;&lt;math&gt;C&lt;/math&gt;&quot;, (2.4, 3.6), SE); label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, (0, 5), N);[/asy] What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36&lt;/math&gt; <br /> <br /> ==Solution==<br /> We can see a Pythagorean triple's two longer lengths: 12, 13. So BD should be 5. This is certainly the case because &lt;math&gt;3^2 + 4^2 = 5^2&lt;/math&gt;, which is &lt;math&gt;BD&lt;/math&gt;. Thus the area of triangle ABD is &lt;math&gt;30&lt;/math&gt;. So &lt;math&gt;30 - 6 = 24&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=88469 2017 AMC 8 Problems/Problem 9 2017-11-22T19:23:51Z <p>StellarG: /* See Also */</p> <hr /> <div>==Problem 9==<br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The 6 green marbles are part of &lt;math&gt;1 - \frac13 - \frac14 = \frac5{12}&lt;/math&gt; of the total marbles. If &lt;math&gt;6 = \frac13&lt;/math&gt; of the total number of marbles, then there would be 18 marbles. Since a fourth of 18 is not a whole number, we cannot have 18 marbles. Then in &lt;math&gt;6 = \frac14&lt;/math&gt; of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 6 - 8 = 10 - 8 = 4 marbles, or D.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=88467 2017 AMC 8 Problems/Problem 19 2017-11-22T19:23:32Z <p>StellarG: Created page with &quot;==Problem 19== For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the lar...&quot;</p> <hr /> <div>==Problem 19==<br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt; <br /> <br /> ==Solution==<br /> Factoring a &lt;math&gt;98!&lt;/math&gt;, we have &lt;math&gt;98!(10,000)&lt;/math&gt;. So we have that &lt;math&gt;98//5 + 98//25 = 19 + 3 = 22&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;. Now &lt;math&gt;10,000&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; factors of &lt;math&gt;5&lt;/math&gt;, so there are &lt;math&gt;22 + 4 = 26&lt;/math&gt;, or &lt;math&gt;D&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=18|num-a=20}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=88463 2017 AMC 8 Problems/Problem 5 2017-11-22T19:20:50Z <p>StellarG: Created page with &quot;==Problem 5== What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;? &lt;math&gt;\textbf{(A) }1020\qqua...&quot;</p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution==<br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;1120&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=20|num-a=22}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=88458 2017 AMC 8 Problems/Problem 9 2017-11-22T19:15:16Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 9==<br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The 6 green marbles are part of &lt;math&gt;1 - \frac13 - \frac14 = \frac5{12}&lt;/math&gt; of the total marbles. If &lt;math&gt;6 = \frac13&lt;/math&gt; of the total number of marbles, then there would be 18 marbles. Since a fourth of 18 is not a whole number, we cannot have 18 marbles. Then in &lt;math&gt;6 = \frac14&lt;/math&gt; of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 6 - 8 = 10 - 8 = 4 marbles, or D.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=20|num-a=22}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=88456 2017 AMC 8 Problems/Problem 9 2017-11-22T19:13:11Z <p>StellarG: /* Problem 9 */</p> <hr /> <div>==Problem 9==<br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The 6 green marbles are part of &lt;math&gt;1 - \frac13 - \frac14 = \frac5{12}&lt;/math&gt; of the total marbles. If &lt;math&gt;6 = \frac13&lt;/math&gt; of the total number of marbles, then there would be 18 marbles. Since a fourth of 18 is not a whole number, we cannot have 18 marbles. Then in &lt;math&gt;6 = \frac14&lt;/math&gt; of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 8 - 8 = 10 - 8 = 2 marbles, or B.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=20|num-a=22}}<br /> <br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=88447 2017 AMC 8 Problems/Problem 9 2017-11-22T19:04:03Z <p>StellarG: Created page with &quot;==Problem 9== All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the sma...&quot;</p> <hr /> <div>==Problem 9==<br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_23&diff=88443 2017 AMC 8 Problems/Problem 23 2017-11-22T19:00:59Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> From the question<br /> &lt;cmath&gt;\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}\in \mathbb{N}&lt;/cmath&gt;<br /> and by simple guess and check, the answer is<br /> &lt;cmath&gt;\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}&lt;/cmath&gt;<br /> -kvedula2004</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_21&diff=88435 2017 AMC 8 Problems/Problem 21 2017-11-22T18:52:34Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem 21==<br /> <br /> Suppose &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are nonzero real numbers, and &lt;math&gt;a+b+c=0&lt;/math&gt;. What are the possible value(s) for &lt;math&gt;\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> There are &lt;math&gt;2&lt;/math&gt; cases to consider:<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: &lt;math&gt;2&lt;/math&gt; of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are positive and the other is negative. WLOG assume that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive and &lt;math&gt;c&lt;/math&gt; is negative. In this case, we have that &lt;cmath&gt;\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.&lt;/cmath&gt;<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: &lt;math&gt;2&lt;/math&gt; of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are negative and the other is positive. WLOG assume that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are negative and &lt;math&gt;c&lt;/math&gt; is positive. In this case, we have that &lt;cmath&gt;\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.&lt;/cmath&gt;<br /> <br /> In both cases, we get that the given expression equals &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ~nukelauncher</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_20&diff=87112 1997 AHSME Problems/Problem 20 2017-08-19T13:59:48Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Which one of the following integers can be expressed as the sum of &lt;math&gt;100&lt;/math&gt; consecutive positive integers?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1,\!627,\!384,\!950\qquad\textbf{(B)}\ 2,\!345,\!678,\!910\qquad\textbf{(C)}\ 3,\!579,\!111,\!300\qquad\textbf{(D)}\ 4,\!692,\!581,\!470\qquad\textbf{(E)}\ 5,\!815,\!937,\!260 &lt;/math&gt;<br /> <br /> ==Solution==<br /> The sum of the first &lt;math&gt;100&lt;/math&gt; integers is &lt;math&gt;\frac{100\cdot 101}{2} = 5050&lt;/math&gt;.<br /> <br /> If you add an integer &lt;math&gt;k&lt;/math&gt; to each of the &lt;math&gt;100&lt;/math&gt; numbers, you get &lt;math&gt;5050 + 100k&lt;/math&gt;, which is the sum of the numbers from &lt;math&gt;k+1&lt;/math&gt; to &lt;math&gt;k+100&lt;/math&gt;.<br /> <br /> You're only adding multiples of &lt;math&gt;100&lt;/math&gt;, so the last two digits will remain unchanged.<br /> <br /> Thus, the only possible answer is &lt;math&gt;\boxed{A}&lt;/math&gt;, because the last two digits are &lt;math&gt;50&lt;/math&gt;.<br /> <br /> As an aside, if &lt;math&gt;5050 + 100k = 1627384950&lt;/math&gt;, then &lt;math&gt;k = 16273799&lt;/math&gt;, and the numbers added are the integers from &lt;math&gt;16273800&lt;/math&gt; to &lt;math&gt;16273899&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Notice how the sum of 100 consecutive integers is &lt;math&gt;(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)&lt;/math&gt;. <br /> <br /> Cancelling out the constants give us &lt;math&gt;100x + 50&lt;/math&gt;. <br /> <br /> Looking over at the list of possible values, we quickly realise that the only possible solution is &lt;math&gt;\boxed{A}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1997|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_8&diff=87070 2007 AIME I Problems/Problem 8 2017-08-17T13:04:20Z <p>StellarG: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The [[polynomial]] &lt;math&gt;P(x)&lt;/math&gt; is [[cubic polynomial | cubic]]. What is the largest value of &lt;math&gt;k&lt;/math&gt; for which the polynomials &lt;math&gt;Q_1(x) = x^2 + (k-29)x - k&lt;/math&gt; and &lt;math&gt;Q_2(x) = 2x^2+ (2k-43)x + k&lt;/math&gt; are both [[factor]]s of &lt;math&gt;P(x)&lt;/math&gt;?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> We can see that &lt;math&gt;Q_1&lt;/math&gt; and &lt;math&gt;Q_2&lt;/math&gt; must have a [[root]] in common for them to both be [[factor]]s of the same cubic.<br /> <br /> Let this root be &lt;math&gt;a&lt;/math&gt;.<br /> <br /> We then know that &lt;math&gt;a&lt;/math&gt; is a root of<br /> &lt;math&gt;<br /> Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0<br /> &lt;/math&gt;<br /> , so &lt;math&gt;x = \frac{-k}{5}&lt;/math&gt;.<br /> <br /> We then know that &lt;math&gt;\frac{-k}{5}&lt;/math&gt; is a root of &lt;math&gt;Q_{1}&lt;/math&gt; so we get:<br /> &lt;math&gt;<br /> \frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k<br /> &lt;/math&gt;<br /> or &lt;math&gt;k^{2}=30k&lt;/math&gt;, so &lt;math&gt;k=30&lt;/math&gt; is the highest.<br /> <br /> We can trivially check into the original equations to find that &lt;math&gt;k=30&lt;/math&gt; produces a root in common, so the answer is &lt;math&gt;030&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Again, let the common root be &lt;math&gt;a&lt;/math&gt;; let the other two roots be &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. We can write that &lt;math&gt;(x - a)(x - m) = x^2 + (k - 29)x - k&lt;/math&gt; and that &lt;math&gt;2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)&lt;/math&gt;.<br /> <br /> Therefore, we can write four equations (and we have four [[variable]]s), &lt;math&gt;a + m = 29 - k&lt;/math&gt;, &lt;math&gt;a + n = \frac{43}{2} - k&lt;/math&gt;, &lt;math&gt;am = -k&lt;/math&gt;, and &lt;math&gt;an = \frac{k}{2}&lt;/math&gt;. <br /> <br /> The first two equations show that &lt;math&gt;m - n = 29 - \frac{43}{2} = \frac{15}{2}&lt;/math&gt;. The last two equations show that &lt;math&gt;\frac{m}{n} = -2&lt;/math&gt;. Solving these show that &lt;math&gt;m = 5&lt;/math&gt; and that &lt;math&gt;n = -\frac{5}{2}&lt;/math&gt;. Substituting back into the equations, we eventually find that &lt;math&gt;k = 30&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_19&diff=84856 2001 AMC 8 Problems/Problem 19 2017-03-23T16:33:05Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car N's speed and time are shown as solid line, which graph illustrates this?<br /> <br /> &lt;asy&gt;<br /> unitsize(12);<br /> <br /> draw((0,9)--(0,0)--(9,0));<br /> label(&quot;time&quot;,(4.5,0),S);<br /> label(&quot;s&quot;,(0,7),W); label(&quot;p&quot;,(0,6),W); label(&quot;e&quot;,(0,5),W); label(&quot;e&quot;,(0,4),W); label(&quot;d&quot;,(0,3),W);<br /> label(&quot;(A)&quot;,(-1,9),NW);<br /> draw((0,4)--(4,4),dashed); label(&quot;M&quot;,(4,4),E);<br /> draw((0,8)--(4,8),linewidth(1)); label(&quot;N&quot;,(4,8),E);<br /> <br /> draw((15,9)--(15,0)--(24,0));<br /> label(&quot;time&quot;,(19.5,0),S);<br /> label(&quot;s&quot;,(15,7),W); label(&quot;p&quot;,(15,6),W); label(&quot;e&quot;,(15,5),W); label(&quot;e&quot;,(15,4),W); label(&quot;d&quot;,(15,3),W);<br /> label(&quot;(B)&quot;,(14,9),NW);<br /> draw((15,4)--(19,4),dashed); label(&quot;M&quot;,(19,4),E);<br /> draw((15,8)--(23,8),linewidth(1)); label(&quot;N&quot;,(23,8),E);<br /> <br /> draw((30,9)--(30,0)--(39,0));<br /> label(&quot;time&quot;,(34.5,0),S);<br /> label(&quot;s&quot;,(30,7),W); label(&quot;p&quot;,(30,6),W); label(&quot;e&quot;,(30,5),W); label(&quot;e&quot;,(30,4),W); label(&quot;d&quot;,(30,3),W);<br /> label(&quot;(C)&quot;,(29,9),NW);<br /> draw((30,4)--(34,4),dashed); label(&quot;M&quot;,(34,4),E);<br /> draw((30,2)--(34,2),linewidth(1)); label(&quot;N&quot;,(34,2),E);<br /> <br /> draw((0,-6)--(0,-15)--(9,-15));<br /> label(&quot;time&quot;,(4.5,-15),S);<br /> label(&quot;s&quot;,(0,-8),W); label(&quot;p&quot;,(0,-9),W); label(&quot;e&quot;,(0,-10),W); label(&quot;e&quot;,(0,-11),W); label(&quot;d&quot;,(0,-12),W);<br /> label(&quot;(D)&quot;,(-1,-6),NW);<br /> draw((0,-11)--(4,-11),dashed); label(&quot;M&quot;,(4,-11),E);<br /> draw((0,-7)--(2,-7),linewidth(1)); label(&quot;N&quot;,(2,-7),E);<br /> <br /> draw((15,-6)--(15,-15)--(24,-15));<br /> label(&quot;time&quot;,(19.5,-15),S);<br /> label(&quot;s&quot;,(15,-8),W); label(&quot;p&quot;,(15,-9),W); label(&quot;e&quot;,(15,-10),W); label(&quot;e&quot;,(15,-11),W); label(&quot;d&quot;,(15,-12),W);<br /> label(&quot;(E)&quot;,(14,-6),NW);<br /> draw((15,-11)--(19,-11),dashed); label(&quot;M&quot;,(19,-11),E);<br /> draw((15,-13)--(23,-13),linewidth(1)); label(&quot;N&quot;,(23,-13),E);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> Since car N has twice the speed, it must be twice as high on the speed axis. Also, since cars M and N travel at the same distance but car N has twice the speed, car N must take half the time. Therefore, line N must be half the size of line M. Since the speeds are constant, both lines are horizontal. Reviewing the graphs, we see that the only one satisfying these conditions is graph &lt;math&gt; \boxed{\text{D}} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2001|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> StellarG https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_7&diff=83813 2017 AMC 10B Problems/Problem 7 2017-02-16T20:21:47Z <p>StellarG: /* Solution */</p> <hr /> <div>==Problem==<br /> Samia set off on her bicycle to visit her friend, traveling at an average speed of &lt;math&gt;17&lt;/math&gt; kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at &lt;math&gt;5&lt;/math&gt; kilometers per hour. In all it took her &lt;math&gt;44&lt;/math&gt; minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let's call the distance that Samia had to travel in total as &lt;math&gt;2x&lt;/math&gt;, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both &lt;math&gt;\frac{2x}{2}&lt;/math&gt;, or &lt;math&gt;x&lt;/math&gt;.<br /> &lt;cmath&gt;&lt;/cmath&gt;<br /> She bikes at a rate of &lt;math&gt;17&lt;/math&gt; kph, so she travels the distance she bikes in &lt;math&gt;\frac{x}{17}&lt;/math&gt; hours. She walks at a rate of &lt;math&gt;5&lt;/math&gt; kph, so she travels the distance she walks in &lt;math&gt;\frac{x}{5}&lt;/math&gt; hours.<br /> &lt;cmath&gt;&lt;/cmath&gt;<br /> The total time is &lt;math&gt;\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}&lt;/math&gt;. This is equal to &lt;math&gt;\frac{44}{60} = \frac{11}{15}&lt;/math&gt; of an hour. Solving for &lt;math&gt;x&lt;/math&gt;, we have:<br /> &lt;cmath&gt;&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{22x}{85} = \frac{11}{15}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{2x}{85} = \frac{1}{15}&lt;/cmath&gt;<br /> &lt;cmath&gt;30x = 85&lt;/cmath&gt;<br /> &lt;cmath&gt;6x = 17&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \frac{17}{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;&lt;/cmath&gt;<br /> Since &lt;math&gt;x&lt;/math&gt; is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about &lt;math&gt;\boxed{\bold{(C)} 2.8}&lt;/math&gt;.<br /> {{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> StellarG