https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathisdecent&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-31T05:05:10Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_25&diff=133892 2007 AMC 12B Problems/Problem 25 2020-09-20T20:28:23Z <p>Mathisdecent: /* Solution */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;A,B,C,D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located in 3-dimensional space with &lt;math&gt;AB=BC=CD=DE=EA=2&lt;/math&gt; and &lt;math&gt;\angle ABC=\angle CDE=\angle DEA=90^o&lt;/math&gt;. The plane of &lt;math&gt;\triangle ABC&lt;/math&gt; is parallel to &lt;math&gt;\overline{DE}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle BDE&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;A=(0,0,0)&lt;/math&gt;, and &lt;math&gt;B=(2,0,0)&lt;/math&gt;. Since &lt;math&gt;EA=2&lt;/math&gt;, we could let &lt;math&gt;C=(2,0,2)&lt;/math&gt;, &lt;math&gt;D=(2,2,2)&lt;/math&gt;, and &lt;math&gt;E=(2,2,0)&lt;/math&gt;. Now to get back to &lt;math&gt;A&lt;/math&gt; we need another vertex &lt;math&gt;F=(0,2,0)&lt;/math&gt;. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw &lt;math&gt;FA&lt;/math&gt;. Now we can bend these three sides into an equilateral triangle, and the coordinates change: &lt;math&gt;A=(0,0,0)&lt;/math&gt;, &lt;math&gt;B=(2,0,0)&lt;/math&gt;, &lt;math&gt;C=(2,0,2)&lt;/math&gt;, &lt;math&gt;D=(1,\sqrt{3},2)&lt;/math&gt;, and &lt;math&gt;E=(1,\sqrt{3},0)&lt;/math&gt;. Checking for all the requirements, they are all satisfied. Now we find the area of triangle &lt;math&gt;BDE&lt;/math&gt;. The side lengths of this triangle are &lt;math&gt;2, 2, 2\sqrt{2}&lt;/math&gt;, which is an isosceles right triangle. Thus the area of it is &lt;math&gt;\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Similar to solution 1, we allow <br /> &lt;math&gt;A=(0,0,0)&lt;/math&gt;, &lt;math&gt;B=(2,0,0)&lt;/math&gt;, and &lt;math&gt;C=(0,2,0)&lt;/math&gt;. This creates the isosceles right triangle on the plane of &lt;math&gt;z=0&lt;/math&gt;<br /> <br /> Now, note that &lt;math&gt;\angle CDE=\angle DEA=90^o&lt;/math&gt;. This means that there exists some vector &lt;math&gt;DE&lt;/math&gt; parallel to the plane of &lt;math&gt;ABC&lt;/math&gt; that forms two right angles with &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. By definition, this is the cross product of the two vectors &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;. Finding this cross product, we take the determinant of vectors<br /> <br /> &lt;math&gt;AE=&lt;x_1,y_1,z&gt;&lt;/math&gt; and<br /> <br /> &lt;math&gt;CD=&lt;x_2,y_2,z&gt;&lt;/math&gt; *Note that z is constant because the line is parallel to the plane*<br /> <br /> to get &lt;math&gt;(y_1-y_2)zi+(x_1-x_2)zj+(x_1y_2-y_1x_2)k&lt;/math&gt;<br /> <br /> Because there can be no movement in the &lt;math&gt;z&lt;/math&gt; direction, the k unit vector must be zero. Also, because the i unit vector must be orthogonal and also 0. Thus, the vector of line &lt;math&gt;DE&lt;/math&gt; is simply &lt;math&gt;2tj+2k&lt;/math&gt;<br /> <br /> From this, you can figure out that line &lt;math&gt;BE=2&lt;/math&gt;, and the area of &lt;math&gt;BDE=\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC12 box|year=2007|ab=B|num-b=24|after=Last Problem}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Mathisdecent https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_24&diff=117459 2020 AMC 10B Problems/Problem 24 2020-02-08T18:23:52Z <p>Mathisdecent: </p> <hr /> <div>==Problem==<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 /> ==Solution==<br /> (Quick solution if you’re in a hurry)<br /> <br /> First notice that the graphs of &lt;math&gt;(x+1000)/70&lt;/math&gt; and &lt;math&gt;\sqrt[]{n}&lt;/math&gt; intersect at 2 points. Then, notice that &lt;math&gt;(n+1000)/70&lt;/math&gt; must be an integer. This means that n is congruent to &lt;math&gt;50 (mod 70)&lt;/math&gt;. <br /> <br /> For the first intersection, testing the first few values of &lt;math&gt;n&lt;/math&gt; (adding &lt;math&gt;70&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; each time and noticing the left side increases by &lt;math&gt;1&lt;/math&gt; each time) yields &lt;math&gt;n=20&lt;/math&gt; and &lt;math&gt;n=21&lt;/math&gt;.<br /> <br /> For the second intersection, using binary search can narrow down the other cases, being &lt;math&gt;n=47&lt;/math&gt;, &lt;math&gt;n=48&lt;/math&gt;, &lt;math&gt;n=49&lt;/math&gt;, and &lt;math&gt;n=50&lt;/math&gt;. This results in a total of 6 cases, for an answer of &lt;math&gt;\boxed{\textbf{(C) }6}&lt;/math&gt;.<br /> <br /> ~DrJoyo<br /> <br /> ==Solution 2 (Graphing)==<br /> <br /> One intuitive approach to the question is graphing. Obviously, you should know what the graph of the square root function is, and if any function is floored (meaning it is taken to the greatest integer less than a value), a stair-like figure should appear. The other function is simply a line with a slope of &lt;math&gt;1/70&lt;/math&gt;. If you precisely draw out the two regions of the graph where the derivative of the square function nears the derivative of the linear function, you can now deduce that &lt;math&gt;3&lt;/math&gt; values of intersection lay closer to the left side of the stair, and &lt;math&gt;3&lt;/math&gt; values lay closer to the right side of the stair.<br /> <br /> With meticulous graphing, you can realize that the answer is &lt;math&gt;\boxed{\textbf{(C) }6}&lt;/math&gt;.<br /> <br /> A in-depth graph with intersection points is linked below.<br /> https://www.desmos.com/calculator/e5wk9adbuk<br /> ==See Also== <br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Mathisdecent https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=102841 2018 AMC 8 Problems/Problem 22 2019-02-15T04:37:30Z <p>Mathisdecent: /* Solution 3 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AAA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Mathisdecent https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=102838 2018 AMC 8 Problems/Problem 22 2019-02-15T04:35:48Z <p>Mathisdecent: /* Solution 2 */</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AAA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> ==Solution 3==<br /> There is a trivial solution.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Mathisdecent https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_11&diff=102433 2019 AMC 10B Problems/Problem 11 2019-02-14T20:00:34Z <p>Mathisdecent: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?<br /> <br /> &lt;math&gt;\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50&lt;/math&gt;<br /> <br /> ==Solution==<br /> Call the amount of marbles in each jar x, because they are equivalent. Thus, x/10 is the amount of green marbles in 1, and x/9 is the amount of green marbles in 2. x/9+x/10=19x/90, 19x/90=95, and x=450 marbles in each jar. Because the 9/10 is the amount of blue marbles in jar 1, and 8/9 is the amount of blue marbles in jar 2, 9x/10-8x/9=x/90, so there must be 5 more marbles in jar 1 than jar 2. The answer is A<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Mathisdecent https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_5&diff=102423 2019 AMC 10B Problems/Problem 5 2019-02-14T19:54:40Z <p>Mathisdecent: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are reflected across the line &lt;math&gt;y=x&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt;, &lt;math&gt;B'&lt;/math&gt;, and &lt;math&gt;C'&lt;/math&gt;, respectively. Assume that none of the vertices of the triangle lie on the line &lt;math&gt;y=x&lt;/math&gt;. Which of the following statements is not always true?<br /> <br /> &lt;math&gt;\textbf{(A) } &lt;/math&gt; Triangle &lt;math&gt;A'B'C'&lt;/math&gt; lies in the first quadrant.<br /> <br /> &lt;math&gt;\textbf{(B) } &lt;/math&gt; Triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'&lt;/math&gt; have the same area.<br /> <br /> &lt;math&gt;\textbf{(C) } &lt;/math&gt; The slope of line &lt;math&gt;AA'&lt;/math&gt; is &lt;math&gt;-1&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) } &lt;/math&gt; The slopes of lines &lt;math&gt;AA'&lt;/math&gt; and &lt;math&gt;CC'&lt;/math&gt; are the same.<br /> <br /> &lt;math&gt;\textbf{(E) } &lt;/math&gt; Lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;A'B'&lt;/math&gt; are perpendicular to each other.<br /> <br /> ==Solution==<br /> Lets analyze all of the options separately. <br /> A: Clearly A is true, because a coordinate in the first quadrant will have (+,+), and its inverse would also have (+,+)<br /> B: The triangles have the same area, it's the same triangle.<br /> C: If coordinate A has (x,y), then its inverse will have (y,x). (x-y)/(y-x)=-1, so this is true.<br /> D: Likewise, if coordinate A has (x1,y1), and AA' has a slope of -1, then coordinate B, with (x2,y2), will also have a slope of -1. This is true. <br /> E: By process of elimination, this is the answer, but if coordinate A has (x1,y1) and coordinate B has (x2,y2), then their inverses will be (y1,x1), (y2,x2), and it is not necessarily true that (y2-y1)/(x2-x1)=-(y2-y1)/(x2-x1). (Negative inverses of each other). Clearly, the answer is E.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Mathisdecent