https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Penguin+spellcaster&feedformat=atom AoPS Wiki - User contributions [en] 2021-10-25T01:54:42Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=137121 2018 AMC 8 Problems/Problem 23 2020-11-09T08:44:50Z <p>Penguin spellcaster: /* Solutions */</p> <hr /> <div>==Problem==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> We will use constructive counting to solve this. There are &lt;math&gt;2&lt;/math&gt; cases: Either all &lt;math&gt;3&lt;/math&gt; points are adjacent, or exactly &lt;math&gt;2&lt;/math&gt; points are adjacent.<br /> <br /> If all &lt;math&gt;3&lt;/math&gt; points are adjacent, then we have &lt;math&gt;8&lt;/math&gt; choices. If we have exactly &lt;math&gt;2&lt;/math&gt; adjacent points, then we will have &lt;math&gt;8&lt;/math&gt; places to put the adjacent points and also &lt;math&gt;4&lt;/math&gt; places to put the remaining point, so we have &lt;math&gt;8\cdot4&lt;/math&gt; choices. The total amount of choices is &lt;math&gt;{8 \choose 3} = 8\cdot7&lt;/math&gt;.<br /> Thus our answer is &lt;math&gt;\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 2 ===<br /> We can decide &lt;math&gt;2&lt;/math&gt; adjacent points with &lt;math&gt;8&lt;/math&gt; choices. The remaining point will have &lt;math&gt;6&lt;/math&gt; choices. However, we have counted the case with &lt;math&gt;3&lt;/math&gt; adjacent points twice, so we need to subtract this case once. The case with the &lt;math&gt;3&lt;/math&gt; adjacent points has &lt;math&gt;8&lt;/math&gt; arrangements, so our answer is &lt;math&gt;\frac{8\cdot6-8}{{8 \choose 3 }}&lt;/math&gt;&lt;math&gt;=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 3 (Stars and Bars)===<br /> Let &lt;math&gt;1&lt;/math&gt; point of the triangle be fixed at the top. Then, there are &lt;math&gt;{7 \choose 2} = 21&lt;/math&gt; ways to chose the other 2 points. There must be &lt;math&gt;3&lt;/math&gt; spaces in the points and &lt;math&gt;3&lt;/math&gt; points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and &lt;math&gt;2&lt;/math&gt; extra points (k-1) distributed so by the stars and bars formula, &lt;math&gt;{n+k-1 \choose k-1}&lt;/math&gt;, there are &lt;math&gt;{4 \choose 2} = 6&lt;/math&gt; ways to arrange the bars and stars. Thus, the probability is &lt;math&gt;\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}&lt;/math&gt;.<br /> The stars and bars formula might be inaccurate<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=VNflxl7VpL0 - Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=137120 2018 AMC 8 Problems/Problem 23 2020-11-09T08:44:35Z <p>Penguin spellcaster: /* Solution */</p> <hr /> <div>==Problem==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=VNflxl7VpL0 - Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=137119 2018 AMC 8 Problems/Problem 23 2020-11-09T08:43:36Z <p>Penguin spellcaster: I did not edit but I want to ask: Are you sure that is the stars and bars formula? I don't think that is the stars and bars formula</p> <hr /> <div>==Problem==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> We will use constructive counting to solve this. There are &lt;math&gt;2&lt;/math&gt; cases: Either all &lt;math&gt;3&lt;/math&gt; points are adjacent, or exactly &lt;math&gt;2&lt;/math&gt; points are adjacent.<br /> <br /> If all &lt;math&gt;3&lt;/math&gt; points are adjacent, then we have &lt;math&gt;8&lt;/math&gt; choices. If we have exactly &lt;math&gt;2&lt;/math&gt; adjacent points, then we will have &lt;math&gt;8&lt;/math&gt; places to put the adjacent points and also &lt;math&gt;4&lt;/math&gt; places to put the remaining point, so we have &lt;math&gt;8\cdot4&lt;/math&gt; choices. The total amount of choices is &lt;math&gt;{8 \choose 3} = 8\cdot7&lt;/math&gt;.<br /> Thus our answer is &lt;math&gt;\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 2 ===<br /> We can decide &lt;math&gt;2&lt;/math&gt; adjacent points with &lt;math&gt;8&lt;/math&gt; choices. The remaining point will have &lt;math&gt;6&lt;/math&gt; choices. However, we have counted the case with &lt;math&gt;3&lt;/math&gt; adjacent points twice, so we need to subtract this case once. The case with the &lt;math&gt;3&lt;/math&gt; adjacent points has &lt;math&gt;8&lt;/math&gt; arrangements, so our answer is &lt;math&gt;\frac{8\cdot6-8}{{8 \choose 3 }}&lt;/math&gt;&lt;math&gt;=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 3 (Stars and Bars)===<br /> Let &lt;math&gt;1&lt;/math&gt; point of the triangle be fixed at the top. Then, there are &lt;math&gt;{7 \choose 2} = 21&lt;/math&gt; ways to chose the other 2 points. There must be &lt;math&gt;3&lt;/math&gt; spaces in the points and &lt;math&gt;3&lt;/math&gt; points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and &lt;math&gt;2&lt;/math&gt; extra points (k-1) distributed so by the stars and bars formula, &lt;math&gt;{n+k-1 \choose k-1}&lt;/math&gt;, there are &lt;math&gt;{4 \choose 2} = 6&lt;/math&gt; ways to arrange the bars and stars. Thus, the probability is &lt;math&gt;\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}&lt;/math&gt;.<br /> The stars and bars formula might be inaccurate<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=VNflxl7VpL0 - Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_4&diff=136893 2003 AMC 12A Problems/Problem 4 2020-11-07T08:55:52Z <p>Penguin spellcaster: Formatting edit</p> <hr /> <div>{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #4]] and [[2003 AMC 10A Problems|2003 AMC 10A #4]]}}<br /> == Problem ==<br /> It takes Anna &lt;math&gt;30&lt;/math&gt; minutes to walk uphill &lt;math&gt;1&lt;/math&gt; km from her home to school, but it takes her only &lt;math&gt;10&lt;/math&gt; minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since she walked &lt;math&gt;1&lt;/math&gt; km to school and &lt;math&gt;1&lt;/math&gt; km back home, her total distance is &lt;math&gt;1+1=2&lt;/math&gt; km. <br /> <br /> Since she spent &lt;math&gt;30&lt;/math&gt; minutes walking to school and &lt;math&gt;10&lt;/math&gt; minutes walking back home, her total time is &lt;math&gt;30+10=40&lt;/math&gt; minutes = &lt;math&gt;\frac{40}{60}=\frac{2}{3}&lt;/math&gt; hours. <br /> <br /> Therefore her average speed in km/hr is &lt;math&gt;\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The average speed of two speeds that travel the same distance is the [[harmonic mean]] of the speeds, or &lt;math&gt;\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}&lt;/math&gt; (for speeds &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;). Mary's speed going to school is &lt;math&gt;2\,\text{km/hr}&lt;/math&gt;, and her speed coming back is &lt;math&gt;6\,\text{km/hr}&lt;/math&gt;. Plugging the numbers in, we get that the average speed is &lt;math&gt;\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{\mathrm{(A)}\ 3}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=3|num-a=5}}<br /> {{AMC12 box|year=2003|ab=A|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_1&diff=136892 2003 AMC 12A Problems/Problem 1 2020-11-07T08:53:16Z <p>Penguin spellcaster: I added a faster solution as solution 5</p> <hr /> <div>{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #1]] and [[2003 AMC 10A Problems|2003 AMC 10A #1]]}}<br /> == Problem ==<br /> What is the difference between the sum of the first &lt;math&gt;2003&lt;/math&gt; even counting numbers and the sum of the first &lt;math&gt;2003&lt;/math&gt; odd counting numbers? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 &lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> The first &lt;math&gt;2003&lt;/math&gt; even counting numbers are &lt;math&gt;2,4,6,...,4006&lt;/math&gt;. <br /> <br /> The first &lt;math&gt;2003&lt;/math&gt; odd counting numbers are &lt;math&gt;1,3,5,...,4005&lt;/math&gt;. <br /> <br /> Thus, the problem is asking for the value of &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005)&lt;/math&gt;. <br /> <br /> &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) &lt;/math&gt; <br /> <br /> &lt;math&gt;= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Using the sum of an [[arithmetic progression]] formula, we can write this as &lt;math&gt;\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;.<br /> <br /> <br /> <br /> ===Solution 3===<br /> The formula for the sum of the first &lt;math&gt;n&lt;/math&gt; even numbers, is &lt;math&gt;S_E=n^{2}+n&lt;/math&gt;, (E standing for even).<br /> <br /> Sum of first &lt;math&gt;n&lt;/math&gt; odd numbers, is &lt;math&gt;S_O=n^{2}&lt;/math&gt;, (O standing for odd).<br /> <br /> Knowing this, plug &lt;math&gt;2003&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt;, <br /> <br /> &lt;math&gt;S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> In the case that we don't know if &lt;math&gt;0&lt;/math&gt; is considered an even number, we note that it doesn't matter! The sum of odd numbers is &lt;math&gt;O=1+3+5+...+4005&lt;/math&gt;. And the sum of even numbers is either &lt;math&gt;E_1=0+2+4...+4004&lt;/math&gt; or &lt;math&gt;E_2=2+4+6+...+4006&lt;/math&gt;. When compared to the sum of odd numbers, we see that each of the &lt;math&gt;n&lt;/math&gt;th term in the series of even numbers differ by &lt;math&gt;1&lt;/math&gt;. For example, take series &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;E_1&lt;/math&gt;. The first terms are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;0&lt;/math&gt;. Their difference is &lt;math&gt;|1-0|=1&lt;/math&gt;. Similarly, take take series &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;E_2&lt;/math&gt;. The first terms are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. Their difference is &lt;math&gt;|1-2|=1&lt;/math&gt;. Since there are &lt;math&gt;2003&lt;/math&gt; terms in each set, the answer &lt;math&gt;\boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;.<br /> <br /> Solution by franzliszt<br /> <br /> ==Solution 5 (Fastest method)==<br /> We can see that the difference of the first even number and the first odd number is one, the difference between the second even number and the second odd number is one and so on. Then, we get &lt;math&gt;1 * 2003&lt;/math&gt; which is &lt;math&gt;\boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;. <br /> <br /> Solution by Penguin Spellcaster<br /> <br /> == See also ==<br /> {{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}}<br /> {{AMC12 box|year=2003|ab=A|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_25&diff=136705 2002 AMC 10A Problems/Problem 25 2020-11-06T10:51:11Z <p>Penguin spellcaster: /* See also */</p> <hr /> <div>== Problem ==<br /> In [[trapezoid]] &lt;math&gt;ABCD&lt;/math&gt; with bases &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, we have &lt;math&gt;AB = 52&lt;/math&gt;, &lt;math&gt;BC = 12&lt;/math&gt;, &lt;math&gt;CD = 39&lt;/math&gt;, and &lt;math&gt;DA = 5&lt;/math&gt;. The area of &lt;math&gt;ABCD&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> It shouldn't be hard to use [[trigonometry]] to bash this and find the height, but there is a much easier way. Extend &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; to meet at point &lt;math&gt;E&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(250);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);<br /> draw(A--B--C--D--cycle);<br /> draw(D--F--C,dashed);<br /> label(&quot;$$A$$&quot;,A,S);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,W);<br /> label(&quot;$$E$$&quot;,F,N);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,E);<br /> label(&quot;12&quot;,(B+C)/2,WSW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since &lt;math&gt;\overline{AB} || \overline{CD}&lt;/math&gt; we have &lt;math&gt;\triangle AEB \sim \triangle DEC&lt;/math&gt;, with the ratio of [[proportion]]ality being &lt;math&gt;\frac {39}{52} = \frac {3}{4}&lt;/math&gt;. Thus<br /> &lt;cmath&gt;<br /> \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} &amp; \Longrightarrow CE = 36 \\<br /> \frac {DE}{DE + 5} = \frac {3}{4} &amp; \Longrightarrow DE = 15 \end{align*}<br /> &lt;/cmath&gt;<br /> So the sides of &lt;math&gt;\triangle CDE&lt;/math&gt; are &lt;math&gt;15,36,39&lt;/math&gt;, which we recognize to be a &lt;math&gt;5 - 12 - 13&lt;/math&gt; [[right triangle]]. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),<br /> &lt;cmath&gt;<br /> [ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Translate the triangle &lt;math&gt;ADD'&lt;/math&gt; so that &lt;math&gt;DD'&lt;/math&gt; coincides with &lt;math&gt;CC'&lt;/math&gt;. We get the following triangle:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0);<br /> draw(A--B--C--cycle);<br /> draw(C--F,dashed);<br /> label(&quot;$$A'$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,N);<br /> label(&quot;$$C'$$&quot;,F,SE);<br /> label(&quot;5&quot;,(A+C)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The length of &lt;math&gt;A'B&lt;/math&gt; in this triangle is equal to the length of the original &lt;math&gt;AB&lt;/math&gt;, minus the length of &lt;math&gt;CD&lt;/math&gt;.<br /> Thus &lt;math&gt;A'B = 52 - 39 = 13&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;A'BC&lt;/math&gt; is a well-known &lt;math&gt;(5,12,13)&lt;/math&gt; right triangle. Its area is &lt;math&gt;[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30&lt;/math&gt;, and therefore its altitude &lt;math&gt;CC'&lt;/math&gt; is &lt;math&gt;\frac{[A'BC]}{A'B} = \frac{60}{13}&lt;/math&gt;.<br /> <br /> Now the area of the original trapezoid is &lt;math&gt;\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> Draw altitudes from points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;:<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(0.2cm);<br /> defaultpen(0.8);<br /> pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0);<br /> draw(A--B--C--D--cycle);<br /> draw(C--E,dashed);<br /> draw(D--F,dashed);<br /> label(&quot;$$A$$&quot;,A,SW);<br /> label(&quot;$$B$$&quot;,B,S);<br /> label(&quot;$$C$$&quot;,C,NE);<br /> label(&quot;$$D$$&quot;,D,N);<br /> label(&quot;$$D'$$&quot;,F,SSE);<br /> label(&quot;$$C'$$&quot;,E,S);<br /> label(&quot;39&quot;,(C+D)/2,N);<br /> label(&quot;52&quot;,(A+B)/2,S);<br /> label(&quot;5&quot;,(A+D)/2,W);<br /> label(&quot;12&quot;,(B+C)/2,ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Call the length of &lt;math&gt;AD'&lt;/math&gt; to be &lt;math&gt;y&lt;/math&gt;, the length of &lt;math&gt;BC'&lt;/math&gt; to be &lt;math&gt;z&lt;/math&gt;, and the height of the trapezoid to be &lt;math&gt;x&lt;/math&gt;.<br /> By the Pythagorean Theorem, we have:<br /> &lt;cmath&gt;z^2 + x^2 = 144&lt;/cmath&gt;<br /> &lt;cmath&gt;y^2 + x^2 = 25&lt;/cmath&gt;<br /> <br /> Subtracting these two equation yields:<br /> &lt;cmath&gt;z^2-y^2=119 \implies (z+y)(z-y)=119&lt;/cmath&gt;<br /> <br /> We also have: &lt;math&gt;z+y=52-39=13&lt;/math&gt;.<br /> <br /> We can substitute the value of &lt;math&gt;z+y&lt;/math&gt; into the equation we just obtained, so we now have:<br /> <br /> &lt;cmath&gt;(13) (z-y)=119 \implies z-y=\frac{119}{13}&lt;/cmath&gt;.<br /> <br /> We can add the &lt;math&gt;z+y&lt;/math&gt; and the &lt;math&gt;z-y&lt;/math&gt; equation to find the value of &lt;math&gt;z&lt;/math&gt;, which simplifies down to be &lt;math&gt;\frac{144}{13}&lt;/math&gt;. Finally, we can plug in &lt;math&gt;z&lt;/math&gt; and use the Pythagorean theorem to find the height of the trapezoid.<br /> <br /> &lt;cmath&gt;\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}&lt;/cmath&gt;<br /> <br /> Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.<br /> <br /> The median of the trapezoid is &lt;math&gt;\frac{39+52}{2} = \frac{91}{2}&lt;/math&gt;, and multiplying this and the height of the trapezoid gets us:<br /> <br /> &lt;cmath&gt;\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}&lt;/cmath&gt;<br /> <br /> === Solution 4 ===<br /> <br /> We construct a line segment parallel to &lt;math&gt;\overline{AD}&lt;/math&gt; from point &lt;math&gt;C&lt;/math&gt; to line &lt;math&gt;\overline{AB},&lt;/math&gt; and label the intersection of this segment with line &lt;math&gt;\overline{AB}&lt;/math&gt; as point &lt;math&gt;E.&lt;/math&gt; Then quadrilateral &lt;math&gt;AECD&lt;/math&gt; is a parallelogram, so &lt;math&gt;CE=5, AE=39,&lt;/math&gt; and &lt;math&gt;EB=13.&lt;/math&gt; Triangle &lt;math&gt;EBC&lt;/math&gt; is therefore a right triangle, with area &lt;math&gt;\frac12 \cdot 5 \cdot 12 = 30.&lt;/math&gt;<br /> <br /> By continuing to split &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt; into segments of length &lt;math&gt;13,&lt;/math&gt; we can connect these vertices in a &quot;zig-zag,&quot; creating seven congruent right triangles, each with sides &lt;math&gt;5,12,&lt;/math&gt; and &lt;math&gt;13,&lt;/math&gt; and each with area &lt;math&gt;30.&lt;/math&gt; The total area is therefore &lt;math&gt;7 \cdot 30 = \boxed{\textbf{(C)} 210}.&lt;/math&gt;<br /> <br /> === Solution 2 but quicker ===<br /> From Solution &lt;math&gt;2&lt;/math&gt; we know that the the altitude of the trapezoid is &lt;math&gt;\frac{60}{13}&lt;/math&gt; and the triangle's area is &lt;math&gt;30&lt;/math&gt;.<br /> Note that once we remove the triangle we get a rectangle with length &lt;math&gt;39&lt;/math&gt; and height &lt;math&gt;\frac{60}{13}&lt;/math&gt;.<br /> The numbers multiply nicely to get &lt;math&gt;180+30=\boxed{(C) 210}&lt;/math&gt;<br /> -harsha12345<br /> <br /> === Quick Time Trouble Solution 5 ===<br /> <br /> First note how the answer choices are all integers.<br /> The area of the trapezoid is &lt;math&gt;\frac{39+52}{2} * h = \frac{91}{2} h&lt;/math&gt;. So h divides 2. Let &lt;math&gt;x&lt;/math&gt; be &lt;math&gt;2h&lt;/math&gt;. The area is now &lt;math&gt;91x&lt;/math&gt;. <br /> Trying x=1 and x=2 can easily be seen to not work. Those make the only integers possible so now you know x is a fraction. <br /> Since the area is an integer the denominator of x must divide either 13 or 7 since &lt;math&gt;91 = 13*7&lt;/math&gt;.<br /> Seeing how &lt;math&gt;39 = 3*13&lt;/math&gt; and &lt;math&gt;52 = 4*13&lt;/math&gt; assume that the denominator divides 13. Letting &lt;math&gt;y = \frac{x}{13}&lt;/math&gt; the area is now &lt;math&gt;7y&lt;/math&gt;.<br /> Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean h is 4 which we ruled out. <br /> So the answer is &lt;math&gt;\boxed{\textbf{(C)} 210}&lt;/math&gt;. - megateleportingrobots<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|num-b=24|after=-(Last question)|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=136316 2019 AMC 8 Problems/Problem 25 2020-11-01T07:25:20Z <p>Penguin spellcaster: /* Formatting edit */</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 /> ==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 /> [[Without loss of generality]], let's 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 /> ==Videos explaining solution==<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> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=User:Penguin_Spellcaster&diff=136315 User:Penguin Spellcaster 2020-11-01T07:22:41Z <p>Penguin spellcaster: /*My Profile thingy*/</p> <hr /> <div>A person who writes his solutions in AoPS</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=136314 2019 AMC 8 Problems/Problem 23 2020-11-01T07:20:59Z <p>Penguin spellcaster: /* Solution 3 */</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;. - juliankuang (lol im smart)<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 /> ==Video explaining solution== <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> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=136313 2019 AMC 8 Problems/Problem 23 2020-11-01T07:18:06Z <p>Penguin spellcaster: /* Replaces partially guessing part with logical reasoning. */</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;. - juliankuang (lol im smart)<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 Penguin_Spellcaster<br /> <br /> ==Video explaining solution== <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> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_25&diff=136235 2009 AMC 8 Problems/Problem 25 2020-10-31T07:43:18Z <p>Penguin spellcaster: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is &lt;math&gt;\frac{1}{2}&lt;/math&gt; foot from the top face. The second cut is &lt;math&gt;\frac{1}{3}&lt;/math&gt; foot below the first cut, and the third cut is &lt;math&gt;\frac{1}{17}&lt;/math&gt; foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?<br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(1,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> void f(real x) {<br /> draw((0,1,x)--(1,1,x)--(1,0,x));<br /> }<br /> f(d);<br /> f(1/6);<br /> f(1/2);<br /> label(&quot;A&quot;, (1,0,3/4), W);<br /> label(&quot;B&quot;, (1,0,1/3), W);<br /> label(&quot;C&quot;, (1,0,1/6-d/4), W);<br /> label(&quot;D&quot;, (1,0,d/2), W);<br /> label(&quot;1/2&quot;, (1,1,3/4), E);<br /> label(&quot;1/3&quot;, (1,1,1/3), E);<br /> label(&quot;1/17&quot;, (0,1,1/6-d/4), E);&lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(2,8/15,7/15);<br /> int t=0;<br /> void f(real x) {<br /> path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle;<br /> path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle;<br /> path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle;<br /> draw(surface(r), white, nolight);<br /> draw(surface(f), white, nolight);<br /> draw(surface(u), white, nolight);<br /> draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x));<br /> t=t+1;<br /> }<br /> f(d);<br /> f(1/2);<br /> f(1/3);<br /> f(1/17);<br /> label(&quot;D&quot;, (1/2, 1, 0), SE);<br /> label(&quot;A&quot;, (1+1/2, 1, 0), SE);<br /> label(&quot;B&quot;, (2+1/2, 1, 0), SE);<br /> label(&quot;C&quot;, (3+1/2, 1, 0), SE);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The areas of the tops of &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; in the figure formed has sum &lt;math&gt; 1+1+1+1 = 4 &lt;/math&gt; as do the bottoms. Thus, the total so far is &lt;math&gt;8&lt;/math&gt;. Now, one of the sides has an area of one, since it combines all of the heights of &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;, which is &lt;math&gt;1&lt;/math&gt;. The other side is also the same. Thus the total area now is &lt;math&gt;10&lt;/math&gt;. From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like a side of &lt;math&gt;A&lt;/math&gt;, with a surface area of half. From the back, it is the same thing. Thus, the total surface area is &lt;math&gt; 10+\frac{1}{2}+\frac{1}{2}= 11 &lt;/math&gt;, or &lt;math&gt; \boxed{\textbf{(E)}\:11 } &lt;/math&gt;.<br /> <br /> ==Solution 2 (Faster method that relies on the MCQ choices)==<br /> <br /> <br /> The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be &lt;math&gt;\boxed{\textbf{(E)}\:11 }&lt;/math&gt; since it is the only answer greater than 10.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_25&diff=136234 2009 AMC 8 Problems/Problem 25 2020-10-31T07:42:17Z <p>Penguin spellcaster: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is &lt;math&gt;\frac{1}{2}&lt;/math&gt; foot from the top face. The second cut is &lt;math&gt;\frac{1}{3}&lt;/math&gt; foot below the first cut, and the third cut is &lt;math&gt;\frac{1}{17}&lt;/math&gt; foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?<br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(1,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> void f(real x) {<br /> draw((0,1,x)--(1,1,x)--(1,0,x));<br /> }<br /> f(d);<br /> f(1/6);<br /> f(1/2);<br /> label(&quot;A&quot;, (1,0,3/4), W);<br /> label(&quot;B&quot;, (1,0,1/3), W);<br /> label(&quot;C&quot;, (1,0,1/6-d/4), W);<br /> label(&quot;D&quot;, (1,0,d/2), W);<br /> label(&quot;1/2&quot;, (1,1,3/4), E);<br /> label(&quot;1/3&quot;, (1,1,1/3), E);<br /> label(&quot;1/17&quot;, (0,1,1/6-d/4), E);&lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> import three;<br /> real d=11/102;<br /> defaultpen(fontsize(8));<br /> defaultpen(linewidth(0.8));<br /> currentprojection=orthographic(2,8/15,7/15);<br /> int t=0;<br /> void f(real x) {<br /> path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle;<br /> path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle;<br /> path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle;<br /> draw(surface(r), white, nolight);<br /> draw(surface(f), white, nolight);<br /> draw(surface(u), white, nolight);<br /> draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x));<br /> t=t+1;<br /> }<br /> f(d);<br /> f(1/2);<br /> f(1/3);<br /> f(1/17);<br /> label(&quot;D&quot;, (1/2, 1, 0), SE);<br /> label(&quot;A&quot;, (1+1/2, 1, 0), SE);<br /> label(&quot;B&quot;, (2+1/2, 1, 0), SE);<br /> label(&quot;C&quot;, (3+1/2, 1, 0), SE);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The areas of the tops of &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; in the figure formed has sum &lt;math&gt; 1+1+1+1 = 4 &lt;/math&gt; as do the bottoms. Thus, the total so far is &lt;math&gt;8&lt;/math&gt;. Now, one of the sides has an area of one, since it combines all of the heights of &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;, which is &lt;math&gt;1&lt;/math&gt;. The other side is also the same. Thus the total area now is &lt;math&gt;10&lt;/math&gt;. From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like a side of &lt;math&gt;A&lt;/math&gt;, with a surface area of half. From the back, it is the same thing. Thus, the total surface area is &lt;math&gt; 10+\frac{1}{2}+\frac{1}{2}= 11 &lt;/math&gt;, or &lt;math&gt; \boxed{\textbf{(E)}\:11 } &lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> <br /> The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be &lt;math&gt;\boxed{\textbf{(E)}\:11 }&lt;/math&gt; since it is the only answer greater than 10.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_24&diff=136233 2009 AMC 8 Problems/Problem 24 2020-10-31T07:30:49Z <p>Penguin spellcaster: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The letters &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; represent digits. If &lt;math&gt; \begin{tabular}{ccc}&amp;A&amp;B\\ +&amp;C&amp;A\\ \hline &amp;D&amp;A\end{tabular} &lt;/math&gt;and &lt;math&gt; \begin{tabular}{ccc}&amp;A&amp;B\\ -&amp;C&amp;A\\ \hline &amp;&amp;A\end{tabular} &lt;/math&gt;,what digit does &lt;math&gt;D&lt;/math&gt; represent?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Because &lt;math&gt; B+A=A &lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt;.<br /> Next, because &lt;math&gt; B-A=A\implies0-A=A,&lt;/math&gt; we get &lt;math&gt;A=5&lt;/math&gt; as the &quot;0&quot; mentioned above is actually 10 in this case. <br /> <br /> Now we can rewrite &lt;math&gt; \begin{tabular}{ccc}&amp;A&amp;B\\ +&amp;C&amp;A\\ \hline &amp;D&amp;A\end{tabular} &lt;/math&gt; as &lt;math&gt; \begin{tabular}{ccc}&amp;5&amp;0\\ +&amp;C&amp;5\\ \hline &amp;D&amp;5\end{tabular} &lt;/math&gt;. Therefore, &lt;math&gt;D=5+C&lt;/math&gt;<br /> <br /> Finally, &lt;math&gt; A-1-C=0\implies{A=C+1}\implies{C=4} &lt;/math&gt;, So we have &lt;math&gt; D=5+C\implies{D=5+4}=\boxed{\textbf{(E)}\ 9 } &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Penguin spellcaster https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_23&diff=136232 2009 AMC 8 Problems/Problem 23 2020-10-31T07:18:59Z <p>Penguin spellcaster: Solution (Does not need quadratic equation)</p> <hr /> <div>==Problem==<br /> <br /> On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought &lt;math&gt;400&lt;/math&gt; jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If there are &lt;math&gt;x&lt;/math&gt; girls, then there are &lt;math&gt;x+2&lt;/math&gt; boys. She gave each girl &lt;math&gt;x&lt;/math&gt; jellybeans and each boy &lt;math&gt;x+2&lt;/math&gt; jellybeans, for a total of &lt;math&gt;x^2 + (x+2)^2&lt;/math&gt; jellybeans. She gave away &lt;math&gt;400-6=394&lt;/math&gt; jellybeans.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> x^2+(x+2)^2 &amp;= 394\\<br /> x^2+x^2+4x+4 &amp;= 394\\<br /> 2x^2 + 4x &amp;= 390\\<br /> x^2 + 2x &amp;= 195\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> From here, we can see that &lt;math&gt;x = 13&lt;/math&gt; as &lt;math&gt;13^2 + 26 = 195&lt;/math&gt;, so there are &lt;math&gt;13&lt;/math&gt; girls, &lt;math&gt;13+2=15&lt;/math&gt; boys, and &lt;math&gt;13+15=\boxed{\textbf{(B)}\ 28}&lt;/math&gt; students.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2009|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Penguin spellcaster