https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Vedadehhc&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-23T03:46:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&diff=146103 2021 AMC 12A Problems/Problem 22 2021-02-12T17:58:14Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Part 1: solving for c <br /> <br /> Notice that &lt;math&gt;\cos \frac{6\pi}7 = \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt; is the negation of the product of roots by Vieta's formulas<br /> <br /> &lt;math&gt;c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Multiply by &lt;math&gt;8 \sin{\frac{2\pi}{7}}&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Then use sine addition formula backwards:<br /> <br /> &lt;math&gt;2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c = -\frac{1}8&lt;/math&gt;<br /> <br /> <br /> Part 2: starting to solve for b<br /> <br /> &lt;math&gt;b&lt;/math&gt; is the sum of roots two at a time by Vieta's<br /> <br /> &lt;math&gt;b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}&lt;/math&gt;<br /> <br /> By plugging all the parts in we get:<br /> <br /> &lt;math&gt; \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 &lt;/math&gt;<br /> <br /> Which ends up being:<br /> <br /> &lt;math&gt; \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 &lt;/math&gt;<br /> <br /> Which is shown in the next part to equal &lt;math&gt;-\frac{1}2&lt;/math&gt;, so &lt;math&gt;b = -\frac{1}2&lt;/math&gt;<br /> <br /> <br /> Part 3: solving for a and b as the sum of roots<br /> <br /> &lt;math&gt;a&lt;/math&gt; is the negation of the sum of roots<br /> <br /> &lt;math&gt;a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)&lt;/math&gt;<br /> <br /> The real values of the 7th roots of unity are: &lt;math&gt;1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7&lt;/math&gt; and they sum to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If we subtract 1, and condense identical terms, we get:<br /> <br /> &lt;math&gt;2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1&lt;/math&gt;<br /> <br /> Therefore, we have &lt;math&gt;a = -\left(-\frac{1}2\right) = \frac{1}2&lt;/math&gt;<br /> <br /> Finally multiply &lt;math&gt;abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}&lt;/math&gt; or &lt;math&gt;\boxed{D) \frac{1}{32}}&lt;/math&gt;.<br /> <br /> ~Tucker<br /> <br /> == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==<br /> https://youtu.be/Im_WTIK0tss<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&diff=146102 2021 AMC 12A Problems/Problem 22 2021-02-12T17:57:23Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Part 1: solving for c <br /> <br /> Notice that &lt;math&gt;\cos \frac{6\pi}7 = \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt; is the negation of the product of roots by Vieta's formulas<br /> <br /> &lt;math&gt;c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Multiply by &lt;math&gt;8 \sin{\frac{2\pi}{7}}&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Then use sine addition formula backwards:<br /> <br /> &lt;math&gt;2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c = -\frac{1}8&lt;/math&gt;<br /> <br /> <br /> Part 2: starting to solve for b<br /> <br /> &lt;math&gt;b&lt;/math&gt; is the sum of roots two at a time by Vieta's<br /> <br /> &lt;math&gt;b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}&lt;/math&gt;<br /> <br /> By plugging all the parts in we get:<br /> <br /> &lt;math&gt; \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 &lt;/math&gt;<br /> <br /> Which ends up being:<br /> <br /> &lt;math&gt; \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 &lt;/math&gt;<br /> <br /> Which is shown in the next part to equal &lt;math&gt;-\frac{1}2&lt;/math&gt;, so &lt;math&gt;b = -\frac{1}2&lt;/math&gt;<br /> <br /> <br /> Part 3: solving for a and b as the sum of roots<br /> <br /> &lt;math&gt;a&lt;/math&gt; is the negation of the sum of roots<br /> <br /> &lt;math&gt;a = -\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> The real values of the 7th roots of unity are: &lt;math&gt;1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7&lt;/math&gt; and they sum to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If we subtract 1, and condense identical terms, we get:<br /> <br /> &lt;math&gt;2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1&lt;/math&gt;<br /> <br /> Therefore, we have &lt;math&gt;a = -\left(-\frac{1}2\right) = \frac{1}2&lt;/math&gt;<br /> <br /> Finally multiply &lt;math&gt;abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}&lt;/math&gt; or &lt;math&gt;\boxed{D) \frac{1}{32}}&lt;/math&gt;.<br /> <br /> ~Tucker<br /> <br /> == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==<br /> https://youtu.be/Im_WTIK0tss<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&diff=146099 2021 AMC 12A Problems/Problem 22 2021-02-12T17:56:40Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Part 1: solving for c <br /> <br /> Notice that &lt;math&gt;\cos \frac{6\pi}7 = \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt; is the negation of the product of roots by Vieta's formulas<br /> <br /> &lt;math&gt;c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Multiply by &lt;math&gt;8 \sin{\frac{2\pi}{7}}&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Then use sine addition formula backwards:<br /> <br /> &lt;math&gt;2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c = -\frac{1}8&lt;/math&gt;<br /> <br /> <br /> Part 2: starting to solve for b<br /> <br /> &lt;math&gt;b&lt;/math&gt; is the sum of roots two at a time by Vieta's<br /> <br /> &lt;math&gt;b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;\cos \alpha \cos \beta = /frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }2&lt;/math&gt;<br /> <br /> By plugging all the parts in we get:<br /> <br /> &lt;math&gt; \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 &lt;/math&gt;<br /> <br /> Which ends up being:<br /> <br /> &lt;math&gt; \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 &lt;/math&gt;<br /> <br /> Which is shown in the next part to equal &lt;math&gt;-\frac{1}2&lt;/math&gt;, so &lt;math&gt;b = -\frac{1}2&lt;/math&gt;<br /> <br /> <br /> Part 3: solving for a and b as the sum of roots<br /> <br /> &lt;math&gt;a&lt;/math&gt; is the negation of the sum of roots<br /> <br /> &lt;math&gt;a = -\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> The real values of the 7th roots of unity are: &lt;math&gt;1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7&lt;/math&gt; and they sum to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If we subtract 1, and condense identical terms, we get:<br /> <br /> &lt;math&gt;2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1&lt;/math&gt;<br /> <br /> Therefore, we have &lt;math&gt;a = -\left(-\frac{1}2\right) = \frac{1}2&lt;/math&gt;<br /> <br /> Finally multiply &lt;math&gt;abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}&lt;/math&gt; or &lt;math&gt;\boxed{D) \frac{1}{32}}&lt;/math&gt;.<br /> <br /> ~Tucker<br /> <br /> == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==<br /> https://youtu.be/Im_WTIK0tss<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&diff=146098 2021 AMC 12A Problems/Problem 22 2021-02-12T17:55:30Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Part 1: solving for c <br /> <br /> Notice that &lt;math&gt;\cos \frac{6\pi}7 = \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt; is the negation of the product of roots by Vieta's formulas<br /> <br /> &lt;math&gt;c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Multiply by &lt;math&gt;8 \sin{\frac{2\pi}{7}}&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Then use sine addition formula backwards:<br /> <br /> &lt;math&gt;2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c = -\frac{1}8&lt;/math&gt;<br /> <br /> <br /> Part 2: starting to solve for b<br /> <br /> &lt;math&gt;b&lt;/math&gt; is the sum of roots two at a time by Vieta's<br /> <br /> &lt;math&gt;b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;\cos \alpha \cos \beta = /frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }2&lt;/math&gt;<br /> <br /> By plugging all the parts in we get:<br /> <br /> &lt;math&gt; \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 &lt;/math&gt;<br /> <br /> Which ends up being:<br /> <br /> &lt;math&gt; \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 &lt;/math&gt;<br /> <br /> Which is shown in the next part to equal &lt;math&gt;-\frac{1}2&lt;/math&gt;, so &lt;math&gt;b = -\frac{1}2&lt;/math&gt;<br /> <br /> <br /> Part 3: solving for a and b as the sum of roots<br /> <br /> &lt;math&gt;a&lt;/math&gt; is the negation of the sum of roots<br /> <br /> &lt;math&gt;a = -\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> The real values of the 7th roots of unity are: &lt;math&gt;1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7&lt;/math&gt; and they sum to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If we subtract 1, and condense identical terms, we get:<br /> <br /> &lt;math&gt;2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1&lt;/math&gt;<br /> <br /> Therefore, we have &lt;math&gt;a = -\left(-\frac{1}2\right) = \frac{1}2&lt;/math&gt;<br /> <br /> Finally multiply &lt;math&gt;abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}&lt;/math&gt; or &lt;math&gt;\boxed{D) \frac{1}{32}}&lt;/math&gt;.<br /> <br /> ~Tucker<br /> <br /> == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==<br /> https://youtu.be/Im_WTIK0tss<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&diff=146097 2021 AMC 12A Problems/Problem 22 2021-02-12T17:54:57Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Part 1: solving for c <br /> <br /> Notice that &lt;math&gt;\cos \frac{6\pi}7 = \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c&lt;/math&gt; is the negation of the product of roots by Vieta's formulas<br /> <br /> &lt;math&gt;c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Multiply by &lt;math&gt;8 \frac{\sin{2\pi}}{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -8 \frac{\sin{2\pi}}{7} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> Then use sine addition formula backwards:<br /> <br /> &lt;math&gt;2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7&lt;/math&gt;<br /> <br /> &lt;math&gt;c = -\frac{1}8&lt;/math&gt;<br /> <br /> <br /> Part 2: starting to solve for b<br /> <br /> &lt;math&gt;b&lt;/math&gt; is the sum of roots two at a time by Vieta's<br /> <br /> &lt;math&gt;b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> We know that &lt;math&gt;\cos \alpha \cos \beta = /frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }2&lt;/math&gt;<br /> <br /> By plugging all the parts in we get:<br /> <br /> &lt;math&gt; \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 &lt;/math&gt;<br /> <br /> Which ends up being:<br /> <br /> &lt;math&gt; \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 &lt;/math&gt;<br /> <br /> Which is shown in the next part to equal &lt;math&gt;-\frac{1}2&lt;/math&gt;, so &lt;math&gt;b = -\frac{1}2&lt;/math&gt;<br /> <br /> <br /> Part 3: solving for a and b as the sum of roots<br /> <br /> &lt;math&gt;a&lt;/math&gt; is the negation of the sum of roots<br /> <br /> &lt;math&gt;a = -\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7&lt;/math&gt;<br /> <br /> The real values of the 7th roots of unity are: &lt;math&gt;1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7&lt;/math&gt; and they sum to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If we subtract 1, and condense identical terms, we get:<br /> <br /> &lt;math&gt;2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1&lt;/math&gt;<br /> <br /> Therefore, we have &lt;math&gt;a = -\left(-\frac{1}2\right) = \frac{1}2&lt;/math&gt;<br /> <br /> Finally multiply &lt;math&gt;abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}&lt;/math&gt; or &lt;math&gt;\boxed{D) \frac{1}{32}}&lt;/math&gt;.<br /> <br /> ~Tucker<br /> <br /> == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==<br /> https://youtu.be/Im_WTIK0tss<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=Research_Science_Institute&diff=128398 Research Science Institute 2020-07-16T16:37:54Z <p>Vedadehhc: /* The Program */</p> <hr /> <div>The '''Research Science Institute''' (RSI) is a 6-week residential program held each summer at the [[Massachusetts Institute of Technology]] (MIT). <br /> <br /> The dates for the 2008 RSI are June 22 through August 2.<br /> <br /> ==History==<br /> <br /> RSI, originally called the Rickover Science Institute, was founded by US Navy Admiral H. G. Rickover and was first held in 1984 in Northern Virgina. (To this day, alumni of the program are known as &quot;Rickoids&quot;.) RSI's name changed to Research Science Institute in 1987, following Admiral Rickover's death, and after a nomadic existence in the Washington, DC area, RSI moved permanently to MIT in 1992. (Parallel RSI sessions were held at the [[University of California, San Diego]] in 1990 and at the [[California Institute of Technology]] in 2004.)<br /> <br /> RSI is sponsored and run by the [[Center for Excellence in Education]] (CEE), a 501(c)3 nonprofit corporation based in McLean, Virginia. The CEE is the successor of the original Rickover Foundation. <br /> <br /> ==The Program==<br /> <br /> RSI admits approximately 80 students each year, of which about 50 are from the United States. Admission to RSI is extremely competitive. There is no tuition or room-and-board charge to attend RSI. Students in all branches of [[science]] and [[mathematics]] attend the program.<br /> <br /> The core of the program is a four-week [[mentor|mentorship]], in which each student is assigned a [[mentor]] (who may be a [[scientist]], a [[mathematician]], an [[engineer]], a physician, or other person working in a scientific or technical capacity) and performs [[research]] under the supervision of that mentor. At the end of the program, the student presents his/her work in the form of both a written research paper and an oral presentation.<br /> <br /> ==Links==<br /> <br /> * [http://www.cee.org/rsi RSI website] at [http://www.cee.org Center for Excellence in Education]<br /> * [[Science summer programs]]<br /> * [[Mathematics summer programs]]</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_25&diff=116649 2020 AMC 12A Problems/Problem 25 2020-02-02T20:27:33Z <p>Vedadehhc: /* Solution 1 */</p> <hr /> <div>==Problem 25==<br /> The number &lt;math&gt;a=\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers, has the property that the sum of all real numbers &lt;math&gt;x&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \lfloor x \rfloor \cdot \{x\} = a \cdot x^2&lt;/cmath&gt;<br /> is &lt;math&gt;420&lt;/math&gt;, where &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;\{x\}=x- \lfloor x \rfloor&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;p+q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let &lt;math&gt;1&lt;k&lt;2&lt;/math&gt; be the unique solution in this range. Note that &lt;math&gt;ck&lt;/math&gt; is also a solution as long as &lt;math&gt;ck &lt; c+1&lt;/math&gt;, hence all our solutions are &lt;math&gt;k, 2k, ..., bk&lt;/math&gt; for some &lt;math&gt;b&lt;/math&gt;. This sum &lt;math&gt;420&lt;/math&gt; must be between &lt;math&gt;\frac{b(b+1)}{2}&lt;/math&gt; and &lt;math&gt;\frac{(b+1)(b+2)}{2}&lt;/math&gt;, which gives &lt;math&gt;b=28&lt;/math&gt; and &lt;math&gt;k=\frac{420}{406}=\frac{30}{29}&lt;/math&gt;. Plugging this back in gives &lt;math&gt;a=\frac{29 \cdot 1}{30^2} = \frac{29}{900} \implies \boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_7&diff=116648 2020 AMC 10A Problems/Problem 7 2020-02-02T20:19:21Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #5]] and [[2020 AMC 10A Problems|2020 AMC 10A #7]]}}<br /> <br /> ==Problem==<br /> The &lt;math&gt;25&lt;/math&gt; integers from &lt;math&gt;-10&lt;/math&gt; to &lt;math&gt;14,&lt;/math&gt; inclusive, can be arranged to form a &lt;math&gt;5&lt;/math&gt;-by-&lt;math&gt;5&lt;/math&gt; square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> == Solution == <br /> <br /> Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by &lt;math&gt;5&lt;/math&gt; is the total value per row. The sum of the &lt;math&gt;25&lt;/math&gt; integers is &lt;math&gt;-10+9+...+14=11+12+13+14=50&lt;/math&gt;, and the common sum is &lt;math&gt;\frac{50}{5}=\boxed{\text{(C) }10}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 2===<br /> <br /> Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get &lt;math&gt;0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}&lt;/math&gt; as our answer.<br /> ~Baolan<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=6|num-a=8}}<br /> {{AMC12 box|year=2020|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_6&diff=116647 2020 AMC 10A Problems/Problem 6 2020-02-02T20:18:57Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #4]] and [[2020 AMC 10A Problems|2020 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> == Solution ==<br /> The ones digit, for all numbers divisible by 5, must be either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. However, from the restriction in the problem, it must be even, giving us exactly one choice (&lt;math&gt;0&lt;/math&gt;) for this digit. For the middle two digits, we may choose any even integer from &lt;math&gt;[0, 8]&lt;/math&gt;, meaning that we have &lt;math&gt;5&lt;/math&gt; total options. For the first digit, we follow similar intuition but realize that it cannot be &lt;math&gt;0&lt;/math&gt;, hence giving us 4 possibilities. Therefore, using the multiplication rule, we get &lt;math&gt;4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}&lt;/math&gt;. ~ciceronii<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_25&diff=116541 2020 AMC 12A Problems/Problem 25 2020-02-02T03:39:26Z <p>Vedadehhc: /* See Also */</p> <hr /> <div>==Problem 25==<br /> The number &lt;math&gt;a=\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers, has the property that the sum of all real numbers &lt;math&gt;x&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \lfloor x \rfloor \cdot \{x\} = a \cdot x^2&lt;/cmath&gt;<br /> is &lt;math&gt;420&lt;/math&gt;, where &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;\{x\}=x- \lfloor x \rfloor&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;p+q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> ==See Also==<br /> {{AMC12 box|year=2020|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_25&diff=116540 2020 AMC 12A Problems/Problem 25 2020-02-02T03:38:39Z <p>Vedadehhc: </p> <hr /> <div>==Problem 25==<br /> The number &lt;math&gt;a=\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers, has the property that the sum of all real numbers &lt;math&gt;x&lt;/math&gt; satisfying<br /> &lt;cmath&gt; \lfloor x \rfloor \cdot \{x\} = a \cdot x^2&lt;/cmath&gt;<br /> is &lt;math&gt;420&lt;/math&gt;, where &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;\{x\}=x- \lfloor x \rfloor&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;p+q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> ==See Also==<br /> {{AMC12 box|year=2020|ab=A|num-b=24|num-a=26}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_22&diff=116351 2020 AMC 12A Problems/Problem 22 2020-02-01T21:51:15Z <p>Vedadehhc: Created page with &quot;== Problem == Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that &lt;cmath&gt;$(2 + i)^n = a_n + b_ni$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq...&quot;</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Square the given equality to yield<br /> &lt;cmath&gt;<br /> (3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,<br /> &lt;/cmath&gt;<br /> so &lt;math&gt;a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)&lt;/math&gt; and<br /> &lt;cmath&gt;<br /> \sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.<br /> &lt;/cmath&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_23&diff=116350 2020 AMC 10A Problems/Problem 23 2020-02-01T21:49:08Z <p>Vedadehhc: /* See Also */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;(0,0), (4,0),&lt;/math&gt; and &lt;math&gt;(0,3).&lt;/math&gt; Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}, 180^{\circ},&lt;/math&gt; and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> First, any combination of motions we can make must reflect &lt;math&gt;T&lt;/math&gt; an even number of times. This is because every time we reflect &lt;math&gt;T&lt;/math&gt;, it changes orientation. Once &lt;math&gt;T&lt;/math&gt; has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed &lt;math&gt;3&lt;/math&gt; transformations and an even number of them must be reflections, we either reflect &lt;math&gt;T&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; times or &lt;math&gt;2&lt;/math&gt; times.<br /> <br /> <br /> <br /> Case 1: 0 reflections on T<br /> <br /> <br /> <br /> In this case, we must use &lt;math&gt;3&lt;/math&gt; rotations to return &lt;math&gt;T&lt;/math&gt; to its original position. Notice that our set of rotations, &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt;, contains every multiple of &lt;math&gt;90^\circ&lt;/math&gt; except for &lt;math&gt;0^\circ&lt;/math&gt;. We can start with any two rotations &lt;math&gt;a,b&lt;/math&gt; in &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt; and there must be exactly one &lt;math&gt;c \equiv -a - b \pmod{360^\circ}&lt;/math&gt; such that we can use the three rotations &lt;math&gt;(a,b,c)&lt;/math&gt; which ensures that &lt;math&gt;a + b + c \equiv 0^\circ \pmod{360^\circ}&lt;/math&gt;. That way, the composition of rotations &lt;math&gt;a,b,c&lt;/math&gt; yields a full rotation. For example, if &lt;math&gt;a = b = 90^\circ&lt;/math&gt;, then &lt;math&gt;c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}&lt;/math&gt;, so &lt;math&gt;c = 180^\circ&lt;/math&gt; and the rotations &lt;math&gt;(90^\circ,90^\circ,180^\circ)&lt;/math&gt; yields a full rotation.<br /> <br /> The only case in which this fails is when &lt;math&gt;c&lt;/math&gt; would have to equal &lt;math&gt;0^\circ&lt;/math&gt;. This happens when &lt;math&gt;(a,b)&lt;/math&gt; is already a full rotation, namely, &lt;math&gt;(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),&lt;/math&gt; or &lt;math&gt;(270^\circ,90^\circ)&lt;/math&gt;. However, we can simply subtract these three cases from the total. Selecting &lt;math&gt;(a,b)&lt;/math&gt; from &lt;math&gt;\{90^\circ,180^\circ,270^\circ\}&lt;/math&gt; yields &lt;math&gt;3 \cdot 3 = 9&lt;/math&gt; choices, and with &lt;math&gt;3&lt;/math&gt; that fail, we are left with &lt;math&gt;6&lt;/math&gt; combinations for case 1.<br /> <br /> <br /> <br /> Case 2: 2 reflections on T<br /> <br /> <br /> <br /> In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps &lt;math&gt;T&lt;/math&gt; back to itself, inserting a rotation before, between, or after these two reflections would change &lt;math&gt;T&lt;/math&gt;'s final location, meaning that any combination involving two reflections across the x-axis would not map &lt;math&gt;T&lt;/math&gt; back to itself. The same applies to two reflections across the y-axis.<br /> <br /> Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a &lt;math&gt;180^\circ&lt;/math&gt; rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us &lt;math&gt;3! = 6&lt;/math&gt; combinations for case 2.<br /> <br /> Combining both cases we get &lt;math&gt;6 + 6 = \boxed{\textbf{(A) } 12}&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> <br /> https://youtu.be/5TjrDCxTm7Q - &lt;math&gt;Phineas1500&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&diff=116349 2020 AMC 10A Problems/Problem 21 2020-02-01T21:48:51Z <p>Vedadehhc: /* See Also */</p> <hr /> <div>There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First, substitute &lt;math&gt;2^{17}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;. <br /> Then, the given equation becomes &lt;math&gt;\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0&lt;/math&gt;.<br /> Now consider only &lt;math&gt;a^{16}-a^{15}&lt;/math&gt;. This equals &lt;math&gt;a^{15}(a-1)=a^{15}*(2^{17}-1)&lt;/math&gt;.<br /> Note that &lt;math&gt;2^{17}-1&lt;/math&gt; equals &lt;math&gt;2^{16}+2^{15}+...+1&lt;/math&gt;, since the sum of a geometric sequence is &lt;math&gt;\frac{a^n-1}{a-1}&lt;/math&gt;.<br /> Thus, we can see that &lt;math&gt;a^{16}-a^{15}&lt;/math&gt; forms the sum of 17 different powers of 2. <br /> Applying the same method to each of &lt;math&gt;a^{14}-a^{13}&lt;/math&gt;, &lt;math&gt;a^{12}-a^{11}&lt;/math&gt;, ... , &lt;math&gt;a^{2}-a^{1}&lt;/math&gt;, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us &lt;math&gt;17*8=136&lt;/math&gt;.<br /> But we must count also the &lt;math&gt;a^0&lt;/math&gt; term. <br /> Thus, Our answer is &lt;math&gt;136+1=\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~seanyoon777<br /> <br /> == Solution 2 ==<br /> (This is similar to solution 1)<br /> Let &lt;math&gt;x = 2^{17}&lt;/math&gt;. Then, &lt;math&gt;2^{289} = x^{17}&lt;/math&gt;.<br /> The LHS can be rewritten as &lt;math&gt;\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1&lt;/math&gt;. Plugging &lt;math&gt;2^{17}&lt;/math&gt; back in for &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;(2^{17}-1)(2^{15+17}+2^{13+17}+\cdots+2^{1+17})+1=(2^{16}+2^{15}+\cdots+2^{1})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1&lt;/math&gt;. When expanded, this will have &lt;math&gt;17\cdot8+1=137&lt;/math&gt; terms. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> ==Solution 3==<br /> Note that the expression is equal to something slightly lower than &lt;math&gt;2^{272}&lt;/math&gt;. Clearly, answer choices &lt;math&gt;(D)&lt;/math&gt; and &lt;math&gt;(E)&lt;/math&gt; make no sense because the lowest sum for &lt;math&gt;273&lt;/math&gt; terms is &lt;math&gt;2^{273}-1&lt;/math&gt;. &lt;math&gt;(A)&lt;/math&gt; just makes no sense. &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt; are 1 apart, but because the expression is odd, it will have to contain &lt;math&gt;2^0=1&lt;/math&gt;, and because &lt;math&gt;(C)&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; bigger, the answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=Talk:2020_AMC_12A&diff=116296 Talk:2020 AMC 12A 2020-02-01T19:02:25Z <p>Vedadehhc: </p> <hr /> <div>Why is this page locked? Can we get it opened up? - vedadehhc</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116247 2020 AMC 10A Problems/Problem 20 2020-02-01T15:51:31Z <p>Vedadehhc: /* See Also */</p> <hr /> <div>== Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution 1==<br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{D) }360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner)<br /> <br /> ~Ultraman<br /> <br /> ==Solution 2 (Pro Guessing Strats)==<br /> We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to &lt;math&gt;a\sqrt{20}&lt;/math&gt; and &lt;math&gt;b\sqrt{20}&lt;/math&gt;, and because the hypotenuse is 20 we get &lt;math&gt;a+b=20&lt;/math&gt;. Testing small numbers, we get that when &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=18&lt;/math&gt;, &lt;math&gt;ab&lt;/math&gt; is indeed a square. The area of the triangle is thus 60, so the answer is &lt;math&gt;\boxed {\textbf{D) }360}&lt;/math&gt;.<br /> <br /> ~tigershark22<br /> <br /> ==Solution 3 (coordinates)==<br /> &lt;asy&gt;<br /> draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br /> draw(circle((0,0),10));<br /> label(&quot;D&quot;,(10,30),N);<br /> label(&quot;C&quot;,(10,0),E);<br /> label(&quot;B&quot;,(-8,-6),S);<br /> label(&quot;A&quot;,(-10,0),W);<br /> &lt;/asy&gt;<br /> Let the points be &lt;math&gt;A(-10,0)&lt;/math&gt;, &lt;math&gt;\:B(x,y)&lt;/math&gt;, &lt;math&gt;\:C(10,0)&lt;/math&gt;, &lt;math&gt;\:D(10,30)&lt;/math&gt;,and &lt;math&gt;\:E(-5,0)&lt;/math&gt;, respectively. Since &lt;math&gt;B&lt;/math&gt; lies on line &lt;math&gt;DE&lt;/math&gt;, we know that &lt;math&gt;y=2x+10&lt;/math&gt;. Furthermore, since &lt;math&gt;\angle{ABC}=90^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; lies on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, so &lt;math&gt;x^2+y^2=100&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with these equations, we get the solutions &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(-8,-6)&lt;/math&gt;. We immediately discard the &lt;math&gt;(0,0)&lt;/math&gt; solution as &lt;math&gt;y&lt;/math&gt; should be negative. Thus, we conclude that &lt;math&gt;[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=116246 2020 AMC 10A Problems/Problem 13 2020-02-01T15:50:36Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} * 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} * \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} * \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_11&diff=116245 2020 AMC 10A Problems/Problem 11 2020-02-01T15:50:04Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #8]] and [[2020 AMC 10A Problems|2020 AMC 10A #11]]}}<br /> <br /> ==Problem 11==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We can see that &lt;math&gt;44^2&lt;/math&gt; is less than 2020. Therefore, there are &lt;math&gt;1976&lt;/math&gt; of the &lt;math&gt;4040&lt;/math&gt; numbers after &lt;math&gt;2020&lt;/math&gt;. Also, there are &lt;math&gt;2064&lt;/math&gt; numbers that are under and equal to &lt;math&gt;2020&lt;/math&gt;. Since &lt;math&gt;44^2&lt;/math&gt; is &lt;math&gt;1936&lt;/math&gt; it, with the other squares will shift our median's placement up &lt;math&gt;44&lt;/math&gt;. We can find that the median of the whole set is &lt;math&gt;2020.5&lt;/math&gt;, and &lt;math&gt;2020.5-44&lt;/math&gt; gives us &lt;math&gt;1976.5&lt;/math&gt;. Our answer is &lt;math&gt;\boxed{\textbf{(C) } 1976.5}&lt;/math&gt;.<br /> <br /> ~aryam<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{AMC12 box|year=2020|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_10&diff=116244 2020 AMC 10A Problems/Problem 10 2020-02-01T15:49:31Z <p>Vedadehhc: /* See Also */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #7]] and [[2020 AMC 10A Problems|2020 AMC 10A #10]]}}<br /> <br /> ==Problem 10==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The volume of each cube follows the pattern of &lt;math&gt;n^3&lt;/math&gt; ascending, for &lt;math&gt;n&lt;/math&gt; is between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;.<br /> <br /> We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the &lt;math&gt;7\times 7\times 7&lt;/math&gt; cube (which is just &lt;math&gt;7 \times 7 = 49&lt;/math&gt;). The sides areas can be measured as the sum &lt;math&gt;4\sum_{n=0}^{7} n^2&lt;/math&gt;, giving us &lt;math&gt;560&lt;/math&gt;. Structurally, if we examine the tower from the top, we see that it really just forms a &lt;math&gt;7\times 7&lt;/math&gt; square of area &lt;math&gt;49&lt;/math&gt;. Therefore, we can say that the total surface area is &lt;math&gt;560 + 49 + 49 = \boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> Alternatively, for the area of the tops, we could have found the sum &lt;math&gt;\sum_{n=0}^{6}((n+1)^{2}-n^{2})&lt;/math&gt;, giving us &lt;math&gt;49&lt;/math&gt; as well.<br /> <br /> ~ciceronii<br /> <br /> ==Solution 2==<br /> <br /> It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.<br /> <br /> First, we will calculate the total surface area of the cubes, ignoring overlap. This value is &lt;math&gt;6\sum_{n=1}^{7} n^2 = 6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6(140) = 840&lt;/math&gt;. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to &lt;math&gt;2\sum_{n=1}^{6} n^2 = 182&lt;/math&gt;. Subtracting the overlapped surface area from the total surface area, we get &lt;math&gt;840 - 182 = \boxed{\textbf{(B) }658}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3 (a bit more tedious than others)==<br /> It can be seen that the side lengths of the cubes using cube roots are all integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt;, inclusive.<br /> <br /> Only the cubes with side length &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; have &lt;math&gt;5&lt;/math&gt; faces in the surface area and the rest have &lt;math&gt;3&lt;/math&gt;. Also, since the <br /> <br /> cubes are stacked, we have to find the difference between each &lt;math&gt;n^2&lt;/math&gt; and &lt;math&gt;(n-1)^2&lt;/math&gt; side length as &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;7&lt;/math&gt; to <br /> <br /> &lt;math&gt;2&lt;/math&gt;. <br /> <br /> We then come up with this: &lt;math&gt;5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)&lt;/math&gt;. <br /> <br /> We then add all of this and get &lt;math&gt;\boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> <br /> ~aryam<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}<br /> {{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_10&diff=116243 2020 AMC 10A Problems/Problem 10 2020-02-01T15:49:12Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #7]] and [[2020 AMC 10A Problems|2020 AMC 10A #10]]}}<br /> <br /> ==Problem 10==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> The volume of each cube follows the pattern of &lt;math&gt;n^3&lt;/math&gt; ascending, for &lt;math&gt;n&lt;/math&gt; is between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;.<br /> <br /> We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the &lt;math&gt;7\times 7\times 7&lt;/math&gt; cube (which is just &lt;math&gt;7 \times 7 = 49&lt;/math&gt;). The sides areas can be measured as the sum &lt;math&gt;4\sum_{n=0}^{7} n^2&lt;/math&gt;, giving us &lt;math&gt;560&lt;/math&gt;. Structurally, if we examine the tower from the top, we see that it really just forms a &lt;math&gt;7\times 7&lt;/math&gt; square of area &lt;math&gt;49&lt;/math&gt;. Therefore, we can say that the total surface area is &lt;math&gt;560 + 49 + 49 = \boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> Alternatively, for the area of the tops, we could have found the sum &lt;math&gt;\sum_{n=0}^{6}((n+1)^{2}-n^{2})&lt;/math&gt;, giving us &lt;math&gt;49&lt;/math&gt; as well.<br /> <br /> ~ciceronii<br /> <br /> ==Solution 2==<br /> <br /> It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.<br /> <br /> First, we will calculate the total surface area of the cubes, ignoring overlap. This value is &lt;math&gt;6\sum_{n=1}^{7} n^2 = 6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6(140) = 840&lt;/math&gt;. Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to &lt;math&gt;2\sum_{n=1}^{6} n^2 = 182&lt;/math&gt;. Subtracting the overlapped surface area from the total surface area, we get &lt;math&gt;840 - 182 = \boxed{\textbf{(B) }658}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3 (a bit more tedious than others)==<br /> It can be seen that the side lengths of the cubes using cube roots are all integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt;, inclusive.<br /> <br /> Only the cubes with side length &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; have &lt;math&gt;5&lt;/math&gt; faces in the surface area and the rest have &lt;math&gt;3&lt;/math&gt;. Also, since the <br /> <br /> cubes are stacked, we have to find the difference between each &lt;math&gt;n^2&lt;/math&gt; and &lt;math&gt;(n-1)^2&lt;/math&gt; side length as &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;7&lt;/math&gt; to <br /> <br /> &lt;math&gt;2&lt;/math&gt;. <br /> <br /> We then come up with this: &lt;math&gt;5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)&lt;/math&gt;. <br /> <br /> We then add all of this and get &lt;math&gt;\boxed{\textbf{(B) }658}&lt;/math&gt;.<br /> <br /> ~aryam<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_10&diff=116241 2020 AMC 12A Problems/Problem 10 2020-02-01T15:47:54Z <p>Vedadehhc: </p> <hr /> <div>==Problem==<br /> <br /> There is a unique positive integer &lt;math&gt;n&lt;/math&gt; such that&lt;cmath&gt;\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Any logarithm in the form &lt;math&gt;\log_{a^b} c = \frac{1}{b} \log_a c&lt;/math&gt;.<br /> <br /> so &lt;cmath&gt;\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}&lt;/cmath&gt;<br /> <br /> becomes<br /> <br /> &lt;cmath&gt;\log_2({\frac{1}{4}\log_{2}{n}}) = \frac{1}{2}\log_2({\frac{1}{2}\log_2{n}})&lt;/cmath&gt;<br /> <br /> Using &lt;math&gt;\log&lt;/math&gt; property of addition, we can expand the parentheses into<br /> <br /> &lt;cmath&gt;\log_2{(\frac{1}{4})}+\log_2{(\log_{2}{n}}) = \frac{1}{2}(\log_2{(\frac{1}{2})} +\log_{2}{(\log_2{n})})&lt;/cmath&gt;<br /> <br /> Expanding the RHS and simplifying the logs without variables, we have<br /> <br /> &lt;cmath&gt;-2+\log_2{(\log_{2}{n}}) = -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})&lt;/cmath&gt;<br /> <br /> Subtracting &lt;math&gt;\frac{1}{2}(\log_{2}{(\log_2{n})})&lt;/math&gt; from both sides and adding &lt;math&gt;2&lt;/math&gt; to both sides gives us<br /> <br /> &lt;cmath&gt;\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}&lt;/cmath&gt;<br /> <br /> Multiplying by &lt;math&gt;2&lt;/math&gt;, raising the logs to exponents of base &lt;math&gt;2&lt;/math&gt; to get rid of the logs and simplifying gives us<br /> <br /> &lt;cmath&gt;(\log_{2}{(\log_2{n})}) = 3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2^{\log_{2}{(\log_2{n})}} = 2^3&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\log_2{n}=8&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;2^{\log_2{n}}=2^8&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n=256&lt;/cmath&gt;<br /> <br /> Adding the digits together, we have &lt;math&gt;2+5+6=\boxed{\textbf{E) }13}&lt;/math&gt; ~quacker88<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=AMC_12_Problems_and_Solutions&diff=115863 AMC 12 Problems and Solutions 2020-02-01T01:20:28Z <p>Vedadehhc: </p> <hr /> <div>[[AMC 12]] problems and solutions. <br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center&quot;<br /> |-<br /> ! Year || Test A || Test B<br /> |-<br /> | 2020 || [[2020 AMC 12A | AMC 12A]] || [[2020 AMC 12B | AMC 12B]]<br /> |-<br /> | 2019 || [[2019 AMC 12A | AMC 12A]] || [[2019 AMC 12B | AMC 12B]]<br /> |-<br /> | 2018 || [[2018 AMC 12A | AMC 12A]] || [[2018 AMC 12B | AMC 12B]]<br /> |-<br /> | 2017 || [[2017 AMC 12A | AMC 12A]] || [[2017 AMC 12B | AMC 12B]]<br /> |-<br /> | 2016 || [[2016 AMC 12A | AMC 12A]] || [[2016 AMC 12B | AMC 12B]]<br /> |-<br /> | 2015 || [[2015 AMC 12A | AMC 12A]] || [[2015 AMC 12B | AMC 12B]]<br /> |-<br /> | 2014 || [[2014 AMC 12A | AMC 12A]] || [[2014 AMC 12B | AMC 12B]]<br /> |-<br /> | 2013 || [[2013 AMC 12A | AMC 12A]] || [[2013 AMC 12B | AMC 12B]]<br /> |-<br /> | 2012 || [[2012 AMC 12A | AMC 12A]] || [[2012 AMC 12B | AMC 12B]]<br /> |-<br /> | 2011 || [[2011 AMC 12A | AMC 12A]] || [[2011 AMC 12B | AMC 12B]]<br /> |-<br /> | 2010 || [[2010 AMC 12A | AMC 12A]] || [[2010 AMC 12B | AMC 12B]]<br /> |-<br /> | 2009 || [[2009 AMC 12A | AMC 12A]] || [[2009 AMC 12B | AMC 12B]]<br /> |-<br /> | 2008 || [[2008 AMC 12A | AMC 12A]] || [[2008 AMC 12B | AMC 12B]]<br /> |-<br /> | 2007 || [[2007 AMC 12A | AMC 12A]] || [[2007 AMC 12B | AMC 12B]]<br /> |-<br /> | 2006 || [[2006 AMC 12A | AMC 12A]] || [[2006 AMC 12B | AMC 12B]]<br /> |-<br /> | 2005 || [[2005 AMC 12A | AMC 12A]] || [[2005 AMC 12B | AMC 12B]]<br /> |-<br /> | 2004 || [[2004 AMC 12A | AMC 12A]] || [[2004 AMC 12B | AMC 12B]]<br /> |-<br /> | 2003 || [[2003 AMC 12A | AMC 12A]] || [[2003 AMC 12B | AMC 12B]]<br /> |-<br /> | 2002 || [[2002 AMC 12A | AMC 12A]] || [[2002 AMC 12B | AMC 12B]]<br /> |-<br /> | 2001<br /> | colspan=&quot;2&quot;| [[2001 AMC 12 | AMC 12]]<br /> |-<br /> | 2000<br /> | colspan=&quot;2&quot;| [[2000 AMC 12 | AMC 12]]<br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Math Contest Problems]]</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_8&diff=114908 2007 AIME I Problems/Problem 8 2020-01-17T16:51:30Z <p>Vedadehhc: /* 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> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_24&diff=114264 2015 AMC 12A Problems/Problem 24 2020-01-04T16:24:15Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen at random among all rational numbers in the interval &lt;math&gt;[0,2)&lt;/math&gt; that can be written as fractions &lt;math&gt;\frac{n}{d}&lt;/math&gt; where &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers with &lt;math&gt;1 \le d \le 5&lt;/math&gt;. What is the probability that<br /> &lt;cmath&gt;(\text{cos}(a\pi)+i\text{sin}(b\pi))^4&lt;/cmath&gt;<br /> is a real number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;\cos(a\pi) = x&lt;/math&gt; and &lt;math&gt;\sin(b\pi) = y&lt;/math&gt;. Consider the binomial expansion of the expression:<br /> &lt;cmath&gt;x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.&lt;/cmath&gt;<br /> <br /> We notice that the only terms with &lt;math&gt;i&lt;/math&gt; are the second and the fourth terms. Thus for the expression to be a real number, either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt;, or the second term and the fourth term cancel each other out (because in the fourth term, you have &lt;math&gt;i^2 = -1&lt;/math&gt;). <br /> <br /> &lt;math&gt;\text{Case~1:}&lt;/math&gt; Either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;. <br /> <br /> The two &lt;math&gt;\text{a's}&lt;/math&gt; satisfying this are &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; and &lt;math&gt;\tfrac{3}{2}&lt;/math&gt;, and the two &lt;math&gt;\text{b's}&lt;/math&gt; satisfying this are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Because &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; can both be expressed as fractions with a denominator less than or equal to &lt;math&gt;5&lt;/math&gt;, there are a total of &lt;math&gt;20&lt;/math&gt; possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.&lt;/cmath&gt;<br /> <br /> Calculating the total number of sets of &lt;math&gt;(a,b)&lt;/math&gt; results in &lt;math&gt;20 \cdot 20 = 400&lt;/math&gt; sets.<br /> Calculating the total number of invalid sets (sets where &lt;math&gt;a&lt;/math&gt; doesn't equal &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; or &lt;math&gt;\tfrac{3}{2}&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; doesn't equal &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;), resulting in &lt;math&gt;(20-2) \cdot (20-2) = 324&lt;/math&gt;.<br /> <br /> Thus the number of valid sets is &lt;math&gt;76&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Case~2}&lt;/math&gt;: The two terms cancel. <br /> <br /> We then have: <br /> <br /> &lt;cmath&gt;\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).&lt;/cmath&gt;<br /> <br /> So: <br /> <br /> &lt;cmath&gt;\cos^2(a\pi) = \sin^2(b\pi),&lt;/cmath&gt;<br /> <br /> which means for a given value of &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt;, there are &lt;math&gt;4&lt;/math&gt; valid values(one in each quadrant).<br /> <br /> When either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; are equal to &lt;math&gt;1&lt;/math&gt;, however, there are only two corresponding values. We don't count the sets where either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt;, for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\tfrac{1}{5}&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; must be &lt;math&gt;\tfrac{3}{10}&lt;/math&gt;, which we don't have). Thus the total number of sets for this case is &lt;math&gt;4 \cdot 4 + 2 \cdot 2 = 20&lt;/math&gt;.<br /> <br /> Thus, our final answer is &lt;math&gt;\frac{(20 + 76)}{400} = \frac{6}{25}&lt;/math&gt;, which is &lt;math&gt;\boxed{\text{(D)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_24&diff=114263 2015 AMC 12A Problems/Problem 24 2020-01-04T15:58:40Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen at random among all rational numbers in the interval &lt;math&gt;[0,2)&lt;/math&gt; that can be written as fractions &lt;math&gt;\frac{n}{d}&lt;/math&gt; where &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers with &lt;math&gt;1 \le d \le 5&lt;/math&gt;. What is the probability that<br /> &lt;cmath&gt;(\text{cos}(a\pi)+i\text{sin}(b\pi))^4&lt;/cmath&gt;<br /> is a real number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1===<br /> <br /> Let &lt;math&gt;\cos(a\pi) = x&lt;/math&gt; and &lt;math&gt;\sin(b\pi) = y&lt;/math&gt;. Consider the binomial expansion of the expression:<br /> &lt;cmath&gt;x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.&lt;/cmath&gt;<br /> <br /> We notice that the only terms with &lt;math&gt;i&lt;/math&gt; are the second and the fourth terms. Thus for the expression to be a real number, either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt;, or the second term and the fourth term cancel each other out (because in the fourth term, you have &lt;math&gt;i^2 = -1&lt;/math&gt;). <br /> <br /> &lt;math&gt;\text{Case~1:}&lt;/math&gt; Either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;. <br /> <br /> The two &lt;math&gt;\text{a's}&lt;/math&gt; satisfying this are &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; and &lt;math&gt;\tfrac{3}{2}&lt;/math&gt;, and the two &lt;math&gt;\text{b's}&lt;/math&gt; satisfying this are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Because &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; can both be expressed as fractions with a denominator less than or equal to &lt;math&gt;5&lt;/math&gt;, there are a total of &lt;math&gt;20&lt;/math&gt; possible values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},&lt;/cmath&gt; <br /> &lt;cmath&gt;\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.&lt;/cmath&gt;<br /> <br /> Calculating the total number of sets of &lt;math&gt;(a,b)&lt;/math&gt; results in &lt;math&gt;20 \cdot 20 = 400&lt;/math&gt; sets.<br /> Calculating the total number of invalid sets (sets where &lt;math&gt;a&lt;/math&gt; doesn't equal &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; or &lt;math&gt;\tfrac{3}{2}&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; doesn't equal &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;), resulting in &lt;math&gt;(20-2) \cdot (20-2) = 324&lt;/math&gt;.<br /> <br /> Thus the number of valid sets is &lt;math&gt;76&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Case~2}&lt;/math&gt;: The two terms cancel. <br /> <br /> We then have: <br /> <br /> &lt;cmath&gt;\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).&lt;/cmath&gt;<br /> <br /> So: <br /> <br /> &lt;cmath&gt;\cos^2(a\pi) = \sin^2(b\pi),&lt;/cmath&gt;<br /> <br /> which means for a given value of &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt;, there are &lt;math&gt;4&lt;/math&gt; valid values(one in each quadrant).<br /> <br /> When either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; are equal to &lt;math&gt;1&lt;/math&gt;, however, there are only two corresponding values. We don't count the sets where either &lt;math&gt;\cos(a\pi)&lt;/math&gt; or &lt;math&gt;\sin(b\pi)&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt;, for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\tfrac{1}{5}&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; must be &lt;math&gt;\tfrac{3}{10}&lt;/math&gt;, which we don't have). Thus the total number of sets for this case is &lt;math&gt;4 \cdot 4 + 2 \cdot 2 = 20&lt;/math&gt;.<br /> <br /> Thus, our final answer is &lt;math&gt;\frac{(20 + 76)}{400} = \frac{6}{25}&lt;/math&gt;, which is &lt;math&gt;\boxed{\text{(D)}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> (WIP)<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_23&diff=112447 2017 AMC 12B Problems/Problem 23 2019-12-01T23:52:59Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>==Problem 23==<br /> The graph of &lt;math&gt;y=f(x)&lt;/math&gt;, where &lt;math&gt;f(x)&lt;/math&gt; is a polynomial of degree &lt;math&gt;3&lt;/math&gt;, contains points &lt;math&gt;A(2,4)&lt;/math&gt;, &lt;math&gt;B(3,9)&lt;/math&gt;, and &lt;math&gt;C(4,16)&lt;/math&gt;. Lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; intersect the graph again at points &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively, and the sum of the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; is 24. What is &lt;math&gt;f(0)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\quad {-2} \qquad \qquad \textbf{(B)}\quad 0 \qquad\qquad \textbf{(C)}\quad 2 \qquad\qquad \textbf{(D)}\quad \dfrac{24}5 \qquad\qquad\textbf{(E)}\quad 8&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, we can define &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;, which contains points &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;. Now we find that lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; are defined by the equations &lt;math&gt;y = 5x - 6&lt;/math&gt;, &lt;math&gt;y= 6x-8&lt;/math&gt;, and &lt;math&gt;y=7x-12&lt;/math&gt; respectively. Since we want to find the &lt;math&gt;x&lt;/math&gt;-coordinates of the intersections of these lines and &lt;math&gt;f(x)&lt;/math&gt;, we set each of them to &lt;math&gt;f(x)&lt;/math&gt;, and synthetically divide by the solutions we already know exist (eg. if we were looking at line &lt;math&gt;AB&lt;/math&gt;, we would synthetically divide by the solutions &lt;math&gt;x=2&lt;/math&gt; and &lt;math&gt;x=3&lt;/math&gt;, because we already know &lt;math&gt;AB&lt;/math&gt; intersects the graph at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, which have &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;). After completing this process on all three lines, we get that the &lt;math&gt;x&lt;/math&gt;-coordinates of &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; are &lt;math&gt;\frac{4a-1}{a}&lt;/math&gt;, &lt;math&gt;\frac{3a-1}{a}&lt;/math&gt;, and &lt;math&gt;\frac{2a-1}{a}&lt;/math&gt; respectively. Adding these together, we get &lt;math&gt;\frac{9a-3}{a} = 24&lt;/math&gt; which gives us &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;. Substituting this back into the original equation, we get &lt;math&gt;f(x) = -\frac{1}{5}(x-2)(x-3)(x-4) + x^2&lt;/math&gt;, and &lt;math&gt;f(0) = -\frac{1}{5}(-2)(-3)(-4) + 0 = \boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 2==<br /> No need to find the equations for the lines, really. First of all, &lt;math&gt;f(x) = a(x-2)(x-3)(x-4) +x^2&lt;/math&gt;. Let's say the line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;y=bx+c&lt;/math&gt;, and &lt;math&gt;x_1&lt;/math&gt; is the &lt;math&gt;x&lt;/math&gt; coordinate of the third intersection, then &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; are the three roots of &lt;math&gt;f(x) - bx-c&lt;/math&gt;. Apparently the value of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; have no effect on the sum of the 3 roots, because the coefficient of the &lt;math&gt;x^2&lt;/math&gt; term is always &lt;math&gt;9a-1&lt;/math&gt;. So we have, <br /> &lt;cmath&gt; \frac{9a-1}{a} = 2+3 + x_1=3+4+x_2 = 2+4+x_3&lt;/cmath&gt;<br /> Add them up we have <br /> &lt;cmath&gt; 3\frac{9a-1}{a} = 18 + x_1+x_2+x_3 = 18 +24&lt;/cmath&gt;<br /> Solve it, we get &lt;math&gt;a = -\frac{1}{5}&lt;/math&gt;.<br /> &lt;math&gt;\boxed{\textbf{(D)}\frac{24}{5}}&lt;/math&gt;.<br /> <br /> - Mathdummy<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_2&diff=112446 2018 AIME I Problems/Problem 2 2019-12-01T23:52:07Z <p>Vedadehhc: /* Solution */</p> <hr /> <div><br /> <br /> ==Problem==<br /> <br /> The number &lt;math&gt;n&lt;/math&gt; can be written in base &lt;math&gt;14&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{b}\text{ }\underline{c}&lt;/math&gt;, can be written in base &lt;math&gt;15&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{b}&lt;/math&gt;, and can be written in base &lt;math&gt;6&lt;/math&gt; as &lt;math&gt;\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }&lt;/math&gt;, where &lt;math&gt;a &gt; 0&lt;/math&gt;. Find the base-&lt;math&gt;10&lt;/math&gt; representation of &lt;math&gt;n&lt;/math&gt;.<br /> ==Solution==<br /> <br /> We have these equations:<br /> &lt;math&gt;196a+14b+c=225a+15c+b=222a+37c&lt;/math&gt;.<br /> Taking the last two we get &lt;math&gt;3a+b=22c&lt;/math&gt;. Because &lt;math&gt;c \neq 0&lt;/math&gt; otherwise &lt;math&gt;a \ngtr 0&lt;/math&gt;, and &lt;math&gt;a \leq 5&lt;/math&gt;, &lt;math&gt;c=1&lt;/math&gt;.<br /> <br /> Then we know &lt;math&gt;3a+b=22&lt;/math&gt;.<br /> Taking the first two equations we see that &lt;math&gt;29a+14=13b&lt;/math&gt;. Combining the two gives &lt;math&gt;a=4, b=10&lt;/math&gt;. Then we see that &lt;math&gt;222 \times 4+37 \times1=\boxed{925}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> We know that &lt;math&gt;196a+14b+c=225a+15c+b=222a+37c&lt;/math&gt;. Combining the first and third equations give that &lt;math&gt;196a+14b+c=222a+37c&lt;/math&gt;, or &lt;cmath&gt;7b=13a+18c&lt;/cmath&gt;<br /> The second and third gives &lt;math&gt;222a+37c=225a+15c+b&lt;/math&gt;, or &lt;cmath&gt;22c-3a=b &lt;/cmath&gt;&lt;cmath&gt; 154c-21a=7b=13a+18c &lt;/cmath&gt;&lt;cmath&gt; 4c=a&lt;/cmath&gt;<br /> We can have &lt;math&gt;a=4,8,12&lt;/math&gt;, but only &lt;math&gt;a=4&lt;/math&gt; falls within the possible digits of base &lt;math&gt;6&lt;/math&gt;. Thus &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;c=1&lt;/math&gt;, and thus you can find &lt;math&gt;b&lt;/math&gt; which equals &lt;math&gt;10&lt;/math&gt;. Thus, our answer is &lt;math&gt;4\cdot225+1\cdot15+0=\boxed{925}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=112445 2018 AMC 10B Problems/Problem 24 2019-12-01T23:51:36Z <p>Vedadehhc: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be a regular hexagon with side length &lt;math&gt;1&lt;/math&gt;. Denote by &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; the midpoints of sides &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{EF}&lt;/math&gt;, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> M=(0,sqrt(3)/2);<br /> N=(3/4,3sqrt(3)/4);<br /> O=(3/4,sqrt(3)/4);<br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$M$&quot;, M, NW);<br /> label(&quot;$N$&quot;, N, NE);<br /> label(&quot;$O$&quot;, O, SE);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> The desired area (hexagon &lt;math&gt;MPNQOR&lt;/math&gt;) consists of an equilateral triangle (&lt;math&gt;\triangle MNO&lt;/math&gt;) and three right triangles (&lt;math&gt;\triangle MPN&lt;/math&gt;, &lt;math&gt;\triangle NQO&lt;/math&gt;, and &lt;math&gt;\triangle ORM&lt;/math&gt;).<br /> <br /> Notice that &lt;math&gt;\overline {AD}&lt;/math&gt; (not shown) and &lt;math&gt;\overline {BC}&lt;/math&gt; are parallel. &lt;math&gt;\overline {XY}&lt;/math&gt; divides transversals &lt;math&gt;\overline {AB}&lt;/math&gt; and &lt;math&gt;\overline {CD}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. Thus, it must also divide transversal &lt;math&gt;\overline {AC}&lt;/math&gt; and transversal &lt;math&gt;\overline {CO}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. By symmetry, the same applies for &lt;math&gt;\overline {CE}&lt;/math&gt; and &lt;math&gt;\overline {EA}&lt;/math&gt; as well as &lt;math&gt;\overline {EM}&lt;/math&gt; and &lt;math&gt;\overline {AN}&lt;/math&gt;.<br /> <br /> <br /> In &lt;math&gt;\triangle ACE&lt;/math&gt;, we see that &lt;math&gt;\frac{[MNO]}{[ACE]} = \frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{[MPN]}{[ACE]} = \frac{1}{8}&lt;/math&gt;. Our desired area becomes <br /> <br /> &lt;cmath&gt;(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Alternate Geometrical Approach to 1)==<br /> Instead of directly finding the desired hexagonal area, &lt;math&gt;\triangle XYZ&lt;/math&gt; can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. See that &lt;math&gt;\triangle XYZ&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equilateral, so &lt;math&gt;m\angle PXN=60&lt;/math&gt;, so &lt;math&gt;m\angle AXP = \frac{180-60}{2}=60&lt;/math&gt;. As &lt;math&gt;\overline {AC}&lt;/math&gt; is a transversal running through &lt;math&gt;\overline {FC}&lt;/math&gt; (use your imagination) and &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;m\angle BAC=m\angle FCA = \frac{m\angle ACE}{2}=30&lt;/math&gt;. <br /> <br /> <br /> Then, &lt;math&gt;\triangle APX&lt;/math&gt; is a 30-60-90 triangle. By HL congruence, &lt;math&gt;\triangle APX \cong \triangle PXN&lt;/math&gt;. &lt;math&gt;AX=\frac{1}{2}&lt;/math&gt;. Then, the area of &lt;math&gt;\triangle PXN&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt;. There are three such triangles for a total area of &lt;math&gt;\triangle XYZ&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt{3}}{32}&lt;/math&gt;. Find the side of &lt;math&gt;\triangle XYZ&lt;/math&gt; to be &lt;math&gt;\frac{3}{2}&lt;/math&gt;, so the area is &lt;math&gt;\frac{9\sqrt{3}}{16}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\frac{9sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32} \implies \boxed{C}&lt;/math&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> whoof<br /> <br /> ==Solution 3 ==<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> <br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, SE);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (&lt;math&gt;AXFZ&lt;/math&gt;, &lt;math&gt;XBCY&lt;/math&gt;, and &lt;math&gt;ZYED&lt;/math&gt;), and 3 right triangles, with one right angle on each of &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt;. <br /> Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal, which has length 2) with a height of &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of &lt;math&gt;\frac{5\sqrt{3}}{16}&lt;/math&gt; for a total area of &lt;math&gt;\frac{15\sqrt{3}}{16}.&lt;/math&gt; (Alternatively, we could have calculated the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt; and subtracted the area of &lt;math&gt;\triangle XYZ&lt;/math&gt;, which, as we showed before, had a side length of 3/2). <br /> Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of &lt;math&gt;YC = 1/2.&lt;/math&gt; Using similar triangles we calculate the base to be 1/4 and the height to be &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; giving us an area of &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt; per triangle, and a total area of &lt;math&gt;3\frac{\sqrt{3}}{32}&lt;/math&gt;. Adding the two areas together, we get &lt;math&gt;\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}&lt;/math&gt;. Finding the total area, we get &lt;math&gt;6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}&lt;/math&gt;. Taking the complement, we get &lt;math&gt;\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution 4 (Trig)==<br /> Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br /> To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is &lt;math&gt;120^{\textrm{o}}&lt;/math&gt; and the trapezoid is isosceles, we know that the angle opposite is &lt;math&gt;60^{\textrm{o}}&lt;/math&gt;, and thus the side length of this triangle is &lt;math&gt;1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}&lt;/math&gt;. So the area of this triangle is &lt;math&gt;\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}&lt;/math&gt;<br /> Now let's find the area of the smaller triangles. Notice, triangle &lt;math&gt;ACE&lt;/math&gt; cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then &lt;math&gt;\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}&lt;/math&gt; and the sum of the areas is &lt;math&gt;3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}&lt;/math&gt;<br /> Therefore, the area of the convex hexagon is &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> Dividing &lt;math&gt;\triangle MNO&lt;/math&gt; into two right triangles congruent to &lt;math&gt;\triangle PMN&lt;/math&gt;, we see that &lt;math&gt;[MPNQOR]=\dfrac{5}{8}[ACE]&lt;/math&gt;. Because &lt;math&gt;[ACE] = \dfrac{1}{2}[ABCDEF]&lt;/math&gt;, we have &lt;math&gt;[MPNQOR]=\dfrac{5}{16}[ABCDEF]&lt;/math&gt;. From here, you should be able to tell that the answer will have a factor of &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt; is the only answer that has a factor of &lt;math&gt;5&lt;/math&gt;. However, if you want to actually calculate the area, you would calculate &lt;math&gt;[ABCDEF]&lt;/math&gt; to be &lt;math&gt;6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=yDbn9Mx2myw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_14&diff=112444 2015 AIME II Problems/Problem 14 2019-12-01T23:50:43Z <p>Vedadehhc: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; be real numbers satisfying &lt;math&gt;x^4y^5+y^4x^5=810&lt;/math&gt; and &lt;math&gt;x^3y^6+y^3x^6=945&lt;/math&gt;. Evaluate &lt;math&gt;2x^3+(xy)^3+2y^3&lt;/math&gt;.<br /> <br /> ==Solution==<br /> The expression we want to find is &lt;math&gt;2(x^3+y^3) + x^3y^3&lt;/math&gt;.<br /> <br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x^3+y^3)=945&lt;/math&gt;, respectively. Dividing the latter by the former equation yields &lt;math&gt;\frac{x^2-xy+y^2}{xy} = \frac{945}{810}&lt;/math&gt;. Adding 3 to both sides and simplifying yields &lt;math&gt;\frac{(x+y)^2}{xy} = \frac{25}{6}&lt;/math&gt;. Solving for &lt;math&gt;x+y&lt;/math&gt; and substituting this expression into the first equation yields &lt;math&gt;\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810&lt;/math&gt;. Solving for &lt;math&gt;xy&lt;/math&gt;, we find that &lt;math&gt;xy = 3\sqrt{2}&lt;/math&gt;, so &lt;math&gt;x^3y^3 = 54&lt;/math&gt;. Substituting this into the second equation and solving for &lt;math&gt;x^3+y^3&lt;/math&gt; yields &lt;math&gt;x^3+y^3=\frac{35}{2}&lt;/math&gt;. So, the expression to evaluate is equal to &lt;math&gt;2 \times \frac{35}{2} + 54 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Factor the given equations as &lt;math&gt;x^4y^4(x+y) = 810&lt;/math&gt; and &lt;math&gt;x^3y^3(x+y)(x^2-xy+y^2)=945&lt;/math&gt;, respectively. By the first equation, &lt;math&gt;x+y=\frac{810}{x^4y^4}&lt;/math&gt;. Plugging this in to the second equation and simplifying yields &lt;math&gt;(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}&lt;/math&gt;. Now substitute &lt;math&gt;\frac{x}{y}=a&lt;/math&gt;. Solving the quadratic in &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;a=\frac{x}{y}=\frac{2}{3}&lt;/math&gt; or &lt;math&gt;\frac{3}{2}&lt;/math&gt; As both of the original equations were symmetric in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, WLOG, let &lt;math&gt;\frac{x}{y}=\frac{2}{3}&lt;/math&gt;, so &lt;math&gt;x=\frac{2}{3}y&lt;/math&gt;. Now plugging this in to either one of the equations, we get the solutions &lt;math&gt;y=\frac{3(2^{\frac{2}{3}})}{2}&lt;/math&gt;, &lt;math&gt;x=2^{\frac{2}{3}}&lt;/math&gt;. Now plugging into what we want, we get &lt;math&gt;8+54+27=\boxed{089}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Add three times the first equation to the second equation and factor to get &lt;math&gt;(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375&lt;/math&gt;. Taking the cube root yields &lt;math&gt;xy(x+y)=15&lt;/math&gt;. Noting that the first equation is &lt;math&gt;(xy)^3\cdot(xy(x+y))=810&lt;/math&gt;, we find that &lt;math&gt;(xy)^3=\frac{810}{15}=54&lt;/math&gt;. Plugging this into the second equation and dividing yields &lt;math&gt;x^3+y^3 = \frac{945}{54} = \frac{35}{2}&lt;/math&gt;. Thus the sum required, as noted in Solution 1, is &lt;math&gt;54+\frac{35}{2}\cdot2 = \boxed{089}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> As with the other solutions, factor. But this time, let &lt;math&gt;a=xy&lt;/math&gt; and &lt;math&gt;b=x+y&lt;/math&gt;. Then &lt;math&gt;a^4b=810&lt;/math&gt;. Notice that &lt;math&gt;x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)&lt;/math&gt;. Now, if we divide the second equation by the first one, we get &lt;math&gt;7/6 = \frac{b^2-3a}{a}&lt;/math&gt;; then &lt;math&gt;\frac{b^2}{a}=\frac{25}{6}&lt;/math&gt;. Therefore, &lt;math&gt;a = \frac{6}{25}b^2&lt;/math&gt;. Substituting &lt;math&gt;a&lt;/math&gt; into &lt;math&gt;b&lt;/math&gt; in equation 2 gives us &lt;math&gt;b^3 = \frac{5^3}{2}&lt;/math&gt;; we are looking for &lt;math&gt;2b(b^2-3a)+a^3&lt;/math&gt;. Finding &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;35&lt;/math&gt;. Substituting into the first equation, we get &lt;math&gt;b=54&lt;/math&gt;. Our final answer is &lt;math&gt;35+54=\boxed{089}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2018_USAJMO_Problems/Problem_2&diff=105326 2018 USAJMO Problems/Problem 2 2019-04-17T14:21:35Z <p>Vedadehhc: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be positive real numbers such that &lt;math&gt;a+b+c=4\sqrt{abc}&lt;/math&gt;. Prove that &lt;cmath&gt;2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> WLOG let &lt;math&gt;a \leq b \leq c&lt;/math&gt;. Add &lt;math&gt;2(ab+bc+ca)&lt;/math&gt; to both sides of the inequality and factor to get: &lt;cmath&gt;4(a(a+b+c)+bc) \geq (a+b+c)^2&lt;/cmath&gt; &lt;cmath&gt;\frac{4a\sqrt{abc}+bc}{2} \geq 2\sqrt{a^2b^2c^2}&lt;/cmath&gt;<br /> <br /> The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.<br /> <br /> ==Solution 2==<br /> WLOG let &lt;math&gt;a \leq b \leq c&lt;/math&gt;. Note that the equations are homogeneous, so WLOG let &lt;math&gt;c=1&lt;/math&gt;.<br /> Thus, the inequality now becomes &lt;math&gt;2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1&lt;/math&gt;, which simplifies to &lt;math&gt;4ab + 3 \geq (a-b)^2&lt;/math&gt;.<br /> <br /> Now we will use the condition. Letting &lt;math&gt;x=a+b&lt;/math&gt; and &lt;math&gt;y=a-b&lt;/math&gt;, we have<br /> &lt;math&gt;x+1=\sqrt{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1&lt;/math&gt;.<br /> <br /> Plugging this into the inequality, we have &lt;math&gt;2x+3 \geq \frac{1}{16}(-x^3+13x^2-3x-1) \implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \geq 0&lt;/math&gt;, which is true since &lt;math&gt;x \geq 0&lt;/math&gt;.<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2018|num-b=1|num-a=3}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=Talk:2004_AMC_10A_Problems/Problem_19&diff=105170 Talk:2004 AMC 10A Problems/Problem 19 2019-04-08T00:05:07Z <p>Vedadehhc: </p> <hr /> <div>The solution to this problem is wrong. The height of parallelogram can not be just equal to the height of cylinder, as in that scenario it cannot circle around the cylinder.<br /> <br /> Not sure what you mean, but the solution seems correct. You might be confusing the length of the strip (and the side length of the parallelogram) with the height of the parallelogram.</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=CSS&diff=105089 CSS 2019-04-02T17:57:56Z <p>Vedadehhc: /* Syntax */</p> <hr /> <div>{{CSS}}<br /> According to Wikipedia, Cascading Style Sheets (CSS) is a style sheet language used to describe the presentation semantics (the look and formatting) of a document written in a markup language.<br /> <br /> Rephrased by some AoPS user, CSS is a language that a webpage can be styled with.<br /> <br /> In Art of Problem Solving, CSS of AoPS user's blogs can be modified by the respective owner.<br /> <br /> == Syntax ==<br /> The syntax of CSS involves a selector, which indicates what you are styling; and curly braces, to delineate the style from the selector. The curly braces contain a list of property: value pairs, separated by a semicolon. Here's an example<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> selector{<br /> property : value;<br /> another-property : value;<br /> /* more properties and values follow */<br /> }<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Note that white space, the last semicolon, and comments (text wrapped in /* comment here */), are complete ignored. However, it is good practice to indent your properties and values for readability.<br /> <br /> If we typed:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> body{<br /> background: red;<br /> }<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> that would make the background of the body element (aka. the entire page) red. A properties final value is the last value that was set. For instance if we typed:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> body{<br /> background: red;<br /> /* extra code may be here */<br /> background: blue;<br /> }<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> and<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> body{ background: red; }<br /> /* extra code may be here */<br /> body{ background: blue; }<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In both cases, the background would be blue.<br /> <br /> == Elements ==<br /> <br /> In webpages, each individual item--a paragraph of text, a button, a text box, a link, even invisible things that exist in a webpage--is called an element. In HTML, an element is identified by the existence of a pair of texts called tags written enclosed in comparison operators (for example, &lt; p &gt;) enclosing the element. For example, this paragraph of text is a p element, as it is enclosed in &lt;p&gt; tags. This p element is a child of a much larger element that covers this entire page, called the &lt;body&gt; element. But it's still contained in the largest element in a HTML page, the &lt;html&gt; element.<br /> <br /> == Properties ==<br /> <br /> An element has various properties attached to it. For example, in the example above, the body element has the property background-color, which determines the background color of the body element. Other properties includes color, text-align, font-family, margin, et cetera.<br /> <br /> == Values ==<br /> <br /> A property always have a value. In this case, the original value of the background-color property of the body element is white, as this page is white. By changing it to the color #FFFF00 (yellow), the page, if it contains that snippet of CSS code, will have a yellow background color.<br /> <br /> To determine the value of a color-related property, one can simply use their English names as the values. However, to make more diverse colors, one has to be acquainted with [http://en.wikipedia.org/wiki/hexadecimal hexadecimal numbers].<br /> <br /> == The CSS Box Model ==<br /> <br /> Elements can be set to be displayed as a block, or box. Most elements are like this by default. Some exceptions are text, links, images, and other elements that modify text. Any element that is a block has a content (modified by the properties width and height), what is in the element, a padding, a space around the content with the same background as the content, a border which goes around the padding, and a margin that is transparent and goes around the border. Elements also have an outline, a less well known property similar to border, except it doesn't take up space (the margin doesn't expand to go around it.)<br /> <br /> == A Good CSS ==<br /> <br /> A CSS can be said good by someone and yet bad by someone else. So it's up to someone to make their own CSS. However, a few guidelines of making a CSS:<br /> <br /> 1. Never make the text unreadable, unless it's the intention of the CSS (which might upset people that don't know it's intended to be so). Also don't make CSS that annoys the reader.<br /> <br /> 2. Black-on-white is the most preferred style, although white-on-black (reverse video) can save power on the reader's computer. Don't make a text that has a very contrasting color with its background, like red-on-cyan, as it causes some difficulty on reading such text.<br /> <br /> == External Links ==<br /> [http://www.w3schools.com/css/] W3Schools CSS Tutorial<br /> <br /> == See also ==<br /> <br /> <br /> {{stub}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104700 2019 AIME I Problems/Problem 6 2019-03-19T15:43:11Z <p>Vedadehhc: removed solution 1 - unclear explanation, and very similar to solution 3 (now solution 2)</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution 1 (Trig)==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 3 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Solution 4 (Algebraic Bashing)==<br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt;. We can use the right triangles in the problem to create equations. Let &lt;math&gt;a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,&lt;/math&gt; and &lt;math&gt;g=NC.&lt;/math&gt; We are trying to find &lt;math&gt;d.&lt;/math&gt; We can find &lt;math&gt;7&lt;/math&gt; equations. They are<br /> &lt;cmath&gt;4225+d^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;4225+d^2+16d+64=a^2+2ab+b^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2=c^2,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2=64,&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2+e^2+2ef+f^2=784,&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2+e^2+2ef+f^2=g^2,&lt;/cmath&gt;<br /> and &lt;cmath&gt;g^2+784=a^2+2ab+b^2.&lt;/cmath&gt;<br /> We can subtract the fifth equation from the sixth equation to get &lt;math&gt;a^2-b^2=g^2-784.&lt;/math&gt; We can subtract the fourth equation from the third equation to get &lt;math&gt;a^2-b^2=c^2-64.&lt;/math&gt; Combining these equations gives &lt;math&gt;c^2-64=g^2-784&lt;/math&gt; so &lt;math&gt;g^2=c^2+720.&lt;/math&gt; Substituting this into the seventh equation gives &lt;math&gt;c^2+1504=a^2+2ab+b^2.&lt;/math&gt; Substituting this into the second equation gives &lt;math&gt;4225+d^2+16d+64=c^2+1504&lt;/math&gt;. Subtracting the first equation from this gives &lt;math&gt;16d+64=1504.&lt;/math&gt; Solving this equation, we find that &lt;math&gt;d=\boxed{090}.&lt;/math&gt;<br /> (Solution by DottedCaculator)<br /> <br /> ==Solution 5 (5-second PoP)==<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair K, L, M, NN, X, O;<br /> K=(-sqrt(98^2+65^2)/2, 0);<br /> NN=(sqrt(98^2+65^2)/2, 0);<br /> L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2)));<br /> M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2)));<br /> X=foot(L, K, NN);<br /> O=extension(L, X, K, M);<br /> draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K));<br /> draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed);<br /> <br /> draw(rightanglemark(K, L, NN, 100));<br /> draw(rightanglemark(K, M, NN, 100));<br /> draw(rightanglemark(L, X, NN, 100));<br /> dot(&quot;$K$&quot;, K, SW);<br /> dot(&quot;$L$&quot;, L, unit(L));<br /> dot(&quot;$M$&quot;, M, unit(M));<br /> dot(&quot;$N$&quot;, NN, SE);<br /> dot(&quot;$X$&quot;, X, S);<br /> &lt;/asy&gt;<br /> Notice that &lt;math&gt;KLMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{KN}&lt;/math&gt; and &lt;math&gt;XOMN&lt;/math&gt; is inscribed in the circle with diameter &lt;math&gt;\overline{ON}&lt;/math&gt;. Furthermore, &lt;math&gt;(XLN)&lt;/math&gt; is tangent to &lt;math&gt;\overline{KL}&lt;/math&gt;. Then, &lt;cmath&gt;KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,&lt;/cmath&gt;and &lt;math&gt;MO=KM-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> (Solution by TheUltimate123)<br /> <br /> ==Solution 6 (Alternative PoP)==<br /> <br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> (Diagram by vedadehhc)<br /> <br /> Call the base of the altitude from &lt;math&gt;L&lt;/math&gt; to &lt;math&gt;NK&lt;/math&gt; point &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;PO=x&lt;/math&gt;. Now, we have that &lt;math&gt;KO=\sqrt{64-x^2}&lt;/math&gt; by the Pythagorean Theorem. Once again by Pythagorean, &lt;math&gt;LO=\sqrt{720+x^2}-x&lt;/math&gt;. Using Power of a Point, we have <br /> <br /> &lt;cmath&gt;(KO)(OM)=(LO)(OQ)&lt;/cmath&gt; (&lt;math&gt;Q&lt;/math&gt; is the intersection of &lt;math&gt;OL&lt;/math&gt; with the circle &lt;math&gt;\neq L&lt;/math&gt;)<br /> <br /> &lt;cmath&gt;8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;8(MO)=720&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;MO=\boxed{090}&lt;/cmath&gt;.<br /> <br /> (Solution by RootThreeOverTwo)<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_9&diff=104542 2019 AIME I Problems/Problem 9 2019-03-15T23:11:20Z <p>Vedadehhc: </p> <hr /> <div>==Problem 9==<br /> Let &lt;math&gt;\tau (n)&lt;/math&gt; denote the number of positive integer divisors of &lt;math&gt;n&lt;/math&gt;. Find the sum of the six least positive integers &lt;math&gt;n&lt;/math&gt; that are solutions to &lt;math&gt;\tau (n) + \tau (n+1) = 7&lt;/math&gt;.<br /> <br /> ==Solution==<br /> In order to obtain a sum of 7, we must have:<br /> * either a number with 5 divisors (''a fourth power of a prime'') and a number with 2 divisors (''a prime''), or<br /> * a number with 4 divisors (''a semiprime'') and a number with 3 divisors (''a square of a prime''). (No integer greater than 1 can have fewer than 2 divisors.)<br /> Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square like &lt;math&gt;3^2&lt;/math&gt; with 3 divisors, or a fourth power like &lt;math&gt;2^4&lt;/math&gt; with 5 divisors. We then find the smallest such values by hand.<br /> * &lt;math&gt;2^2&lt;/math&gt; has two possibilities: 3 and 4, or 4 and 5. Neither works.<br /> * &lt;math&gt;3^2&lt;/math&gt; has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.<br /> * &lt;math&gt;2^4&lt;/math&gt; has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.<br /> * &lt;math&gt;5^2&lt;/math&gt; has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.<br /> * &lt;math&gt;7^2&lt;/math&gt; has two possibilities: 48 and 49, or 49 and 50. Neither works.<br /> * &lt;math&gt;3^4&lt;/math&gt; has two possibilities: 80 and 81, or 81 and 82. Neither works.<br /> * &lt;math&gt;11^2&lt;/math&gt; has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.<br /> * &lt;math&gt;13^2&lt;/math&gt; has two possibilities: 168 and 169, or 169 and 170. Neither works.<br /> * &lt;math&gt;17^2&lt;/math&gt; has two possibilities: 288 and 289, or 289 and 290. Neither works.<br /> * &lt;math&gt;19^2&lt;/math&gt; has two possibilities: 360 and 361, or 361 and 362. Only (361,362) works.<br /> Having computed the working possibilities, we take the sum of the corresponding values of &lt;math&gt;n&lt;/math&gt;: &lt;math&gt;8+9+16+25+121+361 = \boxed{540}&lt;/math&gt;. ~Kepy.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=104540 2019 AIME I Problems/Problem 14 2019-03-15T23:08:27Z <p>Vedadehhc: </p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> The problem tells us that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we get &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. This tells us that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of 16. Since we know &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p(1 - \frac{1}{p})&lt;/math&gt; or &lt;math&gt;p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p&lt;/math&gt; must be &lt;math&gt;1 \pmod{16}&lt;/math&gt;. The two smallest primes that are &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt;, but &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution 1===<br /> &lt;math&gt;\phi(p)&lt;/math&gt; is called the &quot;Euler Function&quot; of integer &lt;math&gt;p&lt;/math&gt;.<br /> Eular theorem: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;n&lt;/math&gt; but relatively prime to &lt;math&gt;n&lt;/math&gt;, then we have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)&lt;/cmath&gt; if &lt;math&gt;(a,p)=1&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104539 2019 AIME I Problems/Problem 6 2019-03-15T23:07:42Z <p>Vedadehhc: </p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution 1 (Trig)==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 3 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104538 2019 AIME I Problems/Problem 6 2019-03-15T23:07:20Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Similar triangles)==<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 2 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104537 2019 AIME I Problems/Problem 6 2019-03-15T23:06:32Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let &lt;math&gt;\angle MKN=\alpha&lt;/math&gt; and &lt;math&gt;\angle LNK=\beta&lt;/math&gt;. Note &lt;math&gt;\angle KLP=\beta&lt;/math&gt;. <br /> <br /> Then, &lt;math&gt;KP=28\sin\beta=8\cos\alpha&lt;/math&gt;.<br /> Furthermore, &lt;math&gt;KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha&lt;/math&gt;.<br /> <br /> Dividing the equations gives<br /> &lt;cmath&gt;\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;MK=\frac{MN}{\tan\alpha}=98&lt;/math&gt;, so &lt;math&gt;MO=MK-KO=\boxed{090}&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt; as shown above. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;. Using these similarities we see that<br /> &lt;cmath&gt;\frac{KP}{KL} = \frac{KL}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;\frac{KP}{KO} = \frac{KM}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}&lt;/cmath&gt;<br /> Combining the two equations, we get<br /> &lt;cmath&gt;\frac{8\cdot KM}{KN} = \frac{784}{KN}&lt;/cmath&gt;<br /> &lt;cmath&gt;8 \cdot KM = 28^2&lt;/cmath&gt;<br /> &lt;cmath&gt;KM = 98&lt;/cmath&gt;<br /> Since &lt;math&gt;KM = KO + MO&lt;/math&gt;, we get &lt;math&gt;MO = 98 -8 = \boxed{090}&lt;/math&gt;.<br /> <br /> Solution by vedadehhc<br /> <br /> ==Solution 2 (Similar triangles, orthocenters)==<br /> Extend &lt;math&gt;KL&lt;/math&gt; and &lt;math&gt;NM&lt;/math&gt; past &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; respectively to meet at &lt;math&gt;P&lt;/math&gt;. Let &lt;math&gt;H&lt;/math&gt; be the intersection of diagonals &lt;math&gt;KM&lt;/math&gt; and &lt;math&gt;LN&lt;/math&gt; (this is the orthocenter of &lt;math&gt;\triangle KNP&lt;/math&gt;).<br /> <br /> As &lt;math&gt;\triangle KOL \sim \triangle KHP&lt;/math&gt; (as &lt;math&gt;LO \parallel PH&lt;/math&gt;, using the fact that &lt;math&gt;H&lt;/math&gt; is the orthocenter), we may let &lt;math&gt;OH = 8k&lt;/math&gt; and &lt;math&gt;LP = 28k&lt;/math&gt;.<br /> <br /> Then using similarity with triangles &lt;math&gt;\triangle KLH&lt;/math&gt; and &lt;math&gt;\triangle KMP&lt;/math&gt; we have<br /> <br /> &lt;cmath&gt;\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}&lt;/cmath&gt;<br /> <br /> Cross-multiplying and dividing by &lt;math&gt;4+4k&lt;/math&gt; gives &lt;math&gt;2(8+8k+HM) = 28 \cdot 7 = 196&lt;/math&gt; so &lt;math&gt;MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}&lt;/math&gt;. (Solution by scrabbler94)<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104490 2019 AIME I Problems/Problem 6 2019-03-15T14:54:31Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, SW);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, SE);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt;. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem. Since &lt;math&gt;\angle KPL \cong \angle KLN&lt;/math&gt; and &lt;math&gt;\angle PKL \cong \angle LKN&lt;/math&gt;, &lt;math&gt;\triangle PKL \sim \triangle LKN&lt;/math&gt; by AA similarity. Similarly, &lt;math&gt;\triangle KMN \sim \triangle KPO&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104485 2019 AIME I Problems/Problem 6 2019-03-15T13:51:31Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, W);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, E);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> label(&quot;65&quot;, ((M.x+N.x)/2, (M.y+N.y)/2), NE);<br /> &lt;/asy&gt;<br /> <br /> First, let &lt;math&gt;P&lt;/math&gt; be the intersection of &lt;math&gt;LO&lt;/math&gt; and &lt;math&gt;KN&lt;/math&gt;. Note that &lt;math&gt;m\angle KPL = 90^{\circ}&lt;/math&gt; as given in the problem.<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104484 2019 AIME I Problems/Problem 6 2019-03-15T13:34:57Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, W);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, E);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> label(&quot;28&quot;, scale(1/2)*L, W);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104482 2019 AIME I Problems/Problem 6 2019-03-15T12:53:34Z <p>Vedadehhc: /* Solution (Similar triangles) */</p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(250);<br /> real h = sqrt(98^2+65^2);<br /> real l = sqrt(h^2-28^2);<br /> pair K = (0,0);<br /> pair N = (h, 0);<br /> pair M = ((98^2)/h, (98*65)/h);<br /> pair L = ((28^2)/h, (28*l)/h);<br /> pair P = ((28^2)/h, 0);<br /> pair O = ((28^2)/h, (8*65)/h);<br /> draw(K--L--N);<br /> draw(K--M--N--cycle);<br /> draw(L--M);<br /> label(&quot;K&quot;, K, W);<br /> label(&quot;L&quot;, L, NW);<br /> label(&quot;M&quot;, M, NE);<br /> label(&quot;N&quot;, N, E);<br /> draw(L--P);<br /> label(&quot;P&quot;, P, S);<br /> dot(O);<br /> label(&quot;O&quot;, shift((1,1))*O, NNE);<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_6&diff=104481 2019 AIME I Problems/Problem 6 2019-03-15T12:23:44Z <p>Vedadehhc: </p> <hr /> <div>==Problem 6==<br /> In convex quadrilateral &lt;math&gt;KLMN&lt;/math&gt; side &lt;math&gt;\overline{MN}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{KM}&lt;/math&gt;, side &lt;math&gt;\overline{KL}&lt;/math&gt; is perpendicular to diagonal &lt;math&gt;\overline{LN}&lt;/math&gt;, &lt;math&gt;MN = 65&lt;/math&gt;, and &lt;math&gt;KL = 28&lt;/math&gt;. The line through &lt;math&gt;L&lt;/math&gt; perpendicular to side &lt;math&gt;\overline{KN}&lt;/math&gt; intersects diagonal &lt;math&gt;\overline{KM}&lt;/math&gt; at &lt;math&gt;O&lt;/math&gt; with &lt;math&gt;KO = 8&lt;/math&gt;. Find &lt;math&gt;MO&lt;/math&gt;.<br /> <br /> ==Solution (Similar triangles)==<br /> (writing this, don't edit)<br /> &lt;asy&gt;<br /> size(100);<br /> draw((0,0)--(1,0));<br /> &lt;/asy&gt;<br /> <br /> ==Video Solution==<br /> Video Solution:<br /> https://www.youtube.com/watch?v=0AXF-5SsLc8<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=North_Carolina_MathCounts&diff=104043 North Carolina MathCounts 2019-03-05T13:33:11Z <p>Vedadehhc: /* NC State Team Members (top 56 placement) */</p> <hr /> <div>North Carolina often fields a top 10 team at national [[MathCounts]] competitions. In recent years, [[Harold Reiter]] has organized practice sessions for the national teams from NC and SC at the [[University of North Carolina at Charlotte]]. <br /> <br /> The top four students from the written round at the state competition are selected for the national team. The coach of the winning school team is offered the opportunity to coach the state team at the national competition.<br /> <br /> <br /> == National Team Awards and Placement ==<br /> * 2018 24th<br /> * 2017 24th<br /> * 2016 6th<br /> * 2015 24th<br /> * 2014 17th<br /> * 2013 6th<br /> * 2012 11th<br /> * 2011 4th<br /> * 2010 17th<br /> * 2009 8th<br /> * 2007 8th<br /> * 2006 9th<br /> * 2005 10th<br /> * 2003 5th<br /> * 1998 5th<br /> <br /> * 1989 1st<br /> * 1988 2nd<br /> * 1987 8th<br /> <br /> == NC State Team Champions ==<br /> * 2018 - Smith Middle School, Chapel Hill<br /> * 2017 - Randolph Middle School, Charlotte<br /> * 2016 - Carnage Middle School, Raleigh<br /> * 2015 - Smith Middle School, Chapel Hill<br /> * 2014 - Carnage Middle School, Raleigh<br /> * 2013 - Randolph Middle School, Charlotte<br /> * 2012 - Carnage Middle School, Raleigh<br /> * 2011 - Ligon Middle School, Raleigh<br /> * 2010 - Ligon Middle School, Raleigh<br /> * 2009 - Ligon Middle School, Raleigh<br /> * 2008 - Ligon Middle School, Raleigh<br /> * 2007 - Arendell Parrott Academy, Kinston<br /> * 2006 - Martin Middle School, Raleigh<br /> * 2005 - [http://web2k.wsfcs.k12.nc.us/hanesms/academics/mcounts.htm Hanes Middle School], Winston-Salem<br /> * 2004 - Ligon Middle School, Raleigh<br /> * 2003 - Guy B. Phillips Middle School, Chapel Hill<br /> * 2002 - Martin Middle School, Raleigh<br /> * 2001 - Ligon Middle School, Raleigh<br /> * 2000 - Durham School of Arts, Durham<br /> * 1999 - Martin Middle School, Raleigh<br /> * 1998 - Charlotte Latin, Charlotte<br /> <br /> * 1988 - Guy B. Phillips Middle School, Chapel Hill<br /> * 1987 - Charlotte Latin, Charlotte<br /> <br /> == NC State Team Members (top 56 placement) ==<br /> * 2018 - Michael Ferguson (50), Eric Wu, Leo DeJong, Terence Zeng, Coach: Boyd Blackburn<br /> * 2017 - Timothy Deng (28), Dev Chheda (38), Owen Deen, Kate Ma, Coach: Harold Reiter <br /> * 2016 - Jerry Zhao (13), Benjamin Wu (51), Jeffrey Liu, David Li, Coach: Ann Chapoton-Genna<br /> * 2015 - Hahn Lheem, Brian Lin, Timothy Deng, Samuel Ferguson, Coach: Boyd Blackburn<br /> * 2014 - Zack Lee (23), Matthew Dai (47), Sahil Patel, Emily Wen, Coach: John Armpriester<br /> * 2013 - Nikhil Reddy (15), Angela Deng (40), Lloyd Liu (42), Andrew Zhang, Coach: Michael Pillsbury<br /> * 2012 - Shenghao Liu (32), Angela Deng (46), Matthew Zheng (48), Zack Lee, Coach: John Armpriester<br /> * 2011 - Michael An (29), Peter Luo (36), Franklin Chen (56), and Evan Liang (52), Coach: Kim Hong Li<br /> * 2010 - Sammy Luo (14), Benjamin Taylor (36), Kavi Jain, Yujian Tang, Coach: Jason Batterson<br /> * 2009 - Calvin Deng (4), Jeffrey An (31), Remy Lee (52), Yu Wang, Coach: Rohan Lewis<br /> * 2008 - Calvin Deng (21), Allen Yang (38), Nick Tobey (40), Connie Zhou, Coach: Rohan Lewis<br /> * 2007 - Jenny Chen (35), Gray Symon (37), Brendan Fletcher (44), David Lucia, Coach: Tonya McLawhorn<br /> * 2006 - Eric &quot;Pom&quot; McCabe (22), Brendan Fletcher, Gray Symon (48), Akash Ganapathi, Coach: Linda Harvel (Southern Middle)<br /> * 2005 - Eric &quot;Pom&quot; McCabe (13), John Berman (45), Bryce Taylor, Kevin Lang, Coach: Greg Taylor<br /> * 2004 - Ray Wang (44), Vivek Bhattacharya, Matthew Adams, Kevin Lang, Coach: Cathy Whelan<br /> * 2003 - Arnav Tripathy (6), Mikhail Lavrov (16), Michael Lin, Vivek Bhattacharya, Coach: Marla McCrea<br /> * 2002 - Mikhail Lavrov, Chris Hoersting, Sunny Xi, Amy Wen, Coach: Jeanette Rojeski<br /> * 2001 - Jeremy Diepenbrock, Jeff Tang, Paul Huang, Sunny Xi, Coach: Pat Heald<br /> * 2000 - John Shoun, Larry Wise, Charles Yu, Justin Lo, Coach: Robin Barbour<br /> * 1999 - Morgan Brown, Edward Su, Jonathan Campbell, Don Swartz, Coach: Lucy Kay<br /> * 1998 - Anders Kaseorg (17), Paul Wrayno, Eric Sundstrom (38), Chris Goulette, Coach: Caroline Wolfe<br /> * 1997 - Anders Kaseorg,?,?,?<br /> <br /> * 1989 - Lenny Ng (1),?,?,?<br /> * 1988 - Lenny Ng (14), Matt Carle (6), Stephen London (33), Nate Bronson, Coach: Betty Chase<br /> * 1987 - Ashley Reiter (3), Stephen London (41), Tim Ross (37), Ghene Faulcon, Coach: Caroline Wolfe<br /> * 1986 - Jeff VanderKam (2),?,?,?</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=North_Carolina_MathCounts&diff=104042 North Carolina MathCounts 2019-03-05T13:32:58Z <p>Vedadehhc: /* NC State Team Champions */</p> <hr /> <div>North Carolina often fields a top 10 team at national [[MathCounts]] competitions. In recent years, [[Harold Reiter]] has organized practice sessions for the national teams from NC and SC at the [[University of North Carolina at Charlotte]]. <br /> <br /> The top four students from the written round at the state competition are selected for the national team. The coach of the winning school team is offered the opportunity to coach the state team at the national competition.<br /> <br /> <br /> == National Team Awards and Placement ==<br /> * 2018 24th<br /> * 2017 24th<br /> * 2016 6th<br /> * 2015 24th<br /> * 2014 17th<br /> * 2013 6th<br /> * 2012 11th<br /> * 2011 4th<br /> * 2010 17th<br /> * 2009 8th<br /> * 2007 8th<br /> * 2006 9th<br /> * 2005 10th<br /> * 2003 5th<br /> * 1998 5th<br /> <br /> * 1989 1st<br /> * 1988 2nd<br /> * 1987 8th<br /> <br /> == NC State Team Champions ==<br /> * 2018 - Smith Middle School, Chapel Hill<br /> * 2017 - Randolph Middle School, Charlotte<br /> * 2016 - Carnage Middle School, Raleigh<br /> * 2015 - Smith Middle School, Chapel Hill<br /> * 2014 - Carnage Middle School, Raleigh<br /> * 2013 - Randolph Middle School, Charlotte<br /> * 2012 - Carnage Middle School, Raleigh<br /> * 2011 - Ligon Middle School, Raleigh<br /> * 2010 - Ligon Middle School, Raleigh<br /> * 2009 - Ligon Middle School, Raleigh<br /> * 2008 - Ligon Middle School, Raleigh<br /> * 2007 - Arendell Parrott Academy, Kinston<br /> * 2006 - Martin Middle School, Raleigh<br /> * 2005 - [http://web2k.wsfcs.k12.nc.us/hanesms/academics/mcounts.htm Hanes Middle School], Winston-Salem<br /> * 2004 - Ligon Middle School, Raleigh<br /> * 2003 - Guy B. Phillips Middle School, Chapel Hill<br /> * 2002 - Martin Middle School, Raleigh<br /> * 2001 - Ligon Middle School, Raleigh<br /> * 2000 - Durham School of Arts, Durham<br /> * 1999 - Martin Middle School, Raleigh<br /> * 1998 - Charlotte Latin, Charlotte<br /> <br /> * 1988 - Guy B. Phillips Middle School, Chapel Hill<br /> * 1987 - Charlotte Latin, Charlotte<br /> <br /> == NC State Team Members (top 56 placement) ==<br /> * 2019 - Minseok Eli Park, Rohit Hari, Sukrith Velmineti, Brian Zhang, Coach: Ann Chapoton-Genna<br /> * 2018 - Michael Ferguson (50), Eric Wu, Leo DeJong, Terence Zeng, Coach: Boyd Blackburn<br /> * 2017 - Timothy Deng (28), Dev Chheda (38), Owen Deen, Kate Ma, Coach: Harold Reiter <br /> * 2016 - Jerry Zhao (13), Benjamin Wu (51), Jeffrey Liu, David Li, Coach: Ann Chapoton-Genna<br /> * 2015 - Hahn Lheem, Brian Lin, Timothy Deng, Samuel Ferguson, Coach: Boyd Blackburn<br /> * 2014 - Zack Lee (23), Matthew Dai (47), Sahil Patel, Emily Wen, Coach: John Armpriester<br /> * 2013 - Nikhil Reddy (15), Angela Deng (40), Lloyd Liu (42), Andrew Zhang, Coach: Michael Pillsbury<br /> * 2012 - Shenghao Liu (32), Angela Deng (46), Matthew Zheng (48), Zack Lee, Coach: John Armpriester<br /> * 2011 - Michael An (29), Peter Luo (36), Franklin Chen (56), and Evan Liang (52), Coach: Kim Hong Li<br /> * 2010 - Sammy Luo (14), Benjamin Taylor (36), Kavi Jain, Yujian Tang, Coach: Jason Batterson<br /> * 2009 - Calvin Deng (4), Jeffrey An (31), Remy Lee (52), Yu Wang, Coach: Rohan Lewis<br /> * 2008 - Calvin Deng (21), Allen Yang (38), Nick Tobey (40), Connie Zhou, Coach: Rohan Lewis<br /> * 2007 - Jenny Chen (35), Gray Symon (37), Brendan Fletcher (44), David Lucia, Coach: Tonya McLawhorn<br /> * 2006 - Eric &quot;Pom&quot; McCabe (22), Brendan Fletcher, Gray Symon (48), Akash Ganapathi, Coach: Linda Harvel (Southern Middle)<br /> * 2005 - Eric &quot;Pom&quot; McCabe (13), John Berman (45), Bryce Taylor, Kevin Lang, Coach: Greg Taylor<br /> * 2004 - Ray Wang (44), Vivek Bhattacharya, Matthew Adams, Kevin Lang, Coach: Cathy Whelan<br /> * 2003 - Arnav Tripathy (6), Mikhail Lavrov (16), Michael Lin, Vivek Bhattacharya, Coach: Marla McCrea<br /> * 2002 - Mikhail Lavrov, Chris Hoersting, Sunny Xi, Amy Wen, Coach: Jeanette Rojeski<br /> * 2001 - Jeremy Diepenbrock, Jeff Tang, Paul Huang, Sunny Xi, Coach: Pat Heald<br /> * 2000 - John Shoun, Larry Wise, Charles Yu, Justin Lo, Coach: Robin Barbour<br /> * 1999 - Morgan Brown, Edward Su, Jonathan Campbell, Don Swartz, Coach: Lucy Kay<br /> * 1998 - Anders Kaseorg (17), Paul Wrayno, Eric Sundstrom (38), Chris Goulette, Coach: Caroline Wolfe<br /> * 1997 - Anders Kaseorg,?,?,?<br /> <br /> * 1989 - Lenny Ng (1),?,?,?<br /> * 1988 - Lenny Ng (14), Matt Carle (6), Stephen London (33), Nate Bronson, Coach: Betty Chase<br /> * 1987 - Ashley Reiter (3), Stephen London (41), Tim Ross (37), Ghene Faulcon, Coach: Caroline Wolfe<br /> * 1986 - Jeff VanderKam (2),?,?,?</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=103942 2019 AMC 10B Problems/Problem 25 2019-02-28T19:06:53Z <p>Vedadehhc: removed solution 3 - the same as solution 1, just less formal</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br /> <br /> ==Problem==<br /> <br /> How many sequences of &lt;math&gt;0&lt;/math&gt;s and &lt;math&gt;1&lt;/math&gt;s of length &lt;math&gt;19&lt;/math&gt; are there that begin with a &lt;math&gt;0&lt;/math&gt;, end with a &lt;math&gt;0&lt;/math&gt;, contain no two consecutive &lt;math&gt;0&lt;/math&gt;s, and contain no three consecutive &lt;math&gt;1&lt;/math&gt;s?<br /> <br /> &lt;math&gt;\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75&lt;/math&gt;<br /> <br /> ==Solution 1 (recursion)==<br /> We can deduce, from the given restrictions, that any valid sequence of length &lt;math&gt;n&lt;/math&gt; will start with a &lt;math&gt;0&lt;/math&gt; followed by either &lt;math&gt;10&lt;/math&gt; or &lt;math&gt;110&lt;/math&gt;.<br /> Thus we can define a recursive function &lt;math&gt;f(n) = f(n-3) + f(n-2)&lt;/math&gt;, where &lt;math&gt;f(n)&lt;/math&gt; is the number of valid sequences of length &lt;math&gt;n&lt;/math&gt;.<br /> <br /> This is because for any valid sequence of length &lt;math&gt;n&lt;/math&gt;, you can append either &lt;math&gt;10&lt;/math&gt; or &lt;math&gt;110&lt;/math&gt; and the resulting sequence will still satisfy the given conditions.<br /> <br /> It is easy to find &lt;math&gt;f(5) = 1&lt;/math&gt; and &lt;math&gt;f(6) = 2&lt;/math&gt; by hand, and then by the recursive formula, we have &lt;math&gt;f(19) = \boxed{\textbf{(C) }65}&lt;/math&gt;.<br /> <br /> ==Solution 2 (casework)==<br /> After any particular &lt;math&gt;0&lt;/math&gt;, the next &lt;math&gt;0&lt;/math&gt; in the sequence must appear exactly &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; positions down the line. In this case, we start at position &lt;math&gt;1&lt;/math&gt; and end at position &lt;math&gt;19&lt;/math&gt;, i.e. we move a total of &lt;math&gt;18&lt;/math&gt; positions down the line. Therefore, we must add a series of &lt;math&gt;2&lt;/math&gt;s and &lt;math&gt;3&lt;/math&gt;s to get &lt;math&gt;18&lt;/math&gt;. There are a number of ways to do this:<br /> <br /> '''Case 1''': nine &lt;math&gt;2&lt;/math&gt;s - there is only &lt;math&gt;1&lt;/math&gt; way to arrange them.<br /> <br /> '''Case 2''': two &lt;math&gt;3&lt;/math&gt;s and six &lt;math&gt;2&lt;/math&gt;s - there are &lt;math&gt;{8\choose2} = 28&lt;/math&gt; ways to arrange them.<br /> <br /> '''Case 3''': four &lt;math&gt;3&lt;/math&gt;s and three &lt;math&gt;2&lt;/math&gt;s - there are &lt;math&gt;{7\choose3} = 35&lt;/math&gt; ways to arrange them.<br /> <br /> '''Case 4''': six &lt;math&gt;3&lt;/math&gt;s - there is only &lt;math&gt;1&lt;/math&gt; way to arrange them.<br /> <br /> Summing the four cases gives &lt;math&gt;1+28+35+1=\boxed{\textbf{(C) }65}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://youtu.be/VamT49PjmdI<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_1&diff=103829 2019 AIME I Problems/Problem 1 2019-02-26T00:29:27Z <p>Vedadehhc: </p> <hr /> <div>The 2019 AIME I takes place on March 13, 2019.<br /> <br /> <br /> {{AIME box|year=2019|n=I|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=103813 2019 AMC 10B Problems/Problem 22 2019-02-25T03:46:14Z <p>Vedadehhc: /* Solution */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br /> <br /> ==Problem==<br /> <br /> Raashan, Sylvia, and Ted play the following game. Each starts with &lt;math&gt; \$1&lt;/math&gt;. A bell rings every &lt;math&gt;15&lt;/math&gt; seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives &lt;math&gt;\$1&lt;/math&gt; to that player. What is the probability that after the bell has rung &lt;math&gt;2019&lt;/math&gt; times, each player will have &lt;math&gt;\$1&lt;/math&gt;? (For example, Raashan and Ted may each decide to give &lt;math&gt;\$1&lt;/math&gt; to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have &lt;math&gt;\$0&lt;/math&gt;, Sylvia will have &lt;math&gt;\$2&lt;/math&gt;, and Ted will have &lt;math&gt;\$1&lt;/math&gt;, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their &lt;math&gt; \$1&lt;/math&gt; to, and the holdings will be the same at the end of the second round.)<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> On the first turn, each player starts off with &lt;math&gt;\$1&lt;/math&gt;. Each turn after that, there are only two possibilities: either everyone stays at &lt;math&gt;\$1&lt;/math&gt;, which we will write as &lt;math&gt;(1-1-1)&lt;/math&gt;, or the distribution of money becomes &lt;math&gt;\$2-\$1-\$0&lt;/math&gt; in some order, which we write as &lt;math&gt;(2-1-0)&lt;/math&gt;. We will consider these two states separately.<br /> <br /> In the &lt;math&gt;(1-1-1)&lt;/math&gt; state, each person has two choices for whom to give their dollar to, meaning there are &lt;math&gt;2^3=8&lt;/math&gt; possible ways that the money can be rearranged. Note that there are only two ways that we can reach &lt;math&gt;(1-1-1)&lt;/math&gt; again: <br /> <br /> 1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.<br /> <br /> 2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.<br /> <br /> Thus, the probability of staying in the &lt;math&gt;(1-1-1)&lt;/math&gt; state is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, while the probability of going to the &lt;math&gt;(2-1-0)&lt;/math&gt; state is &lt;math&gt;\frac{3}{4}&lt;/math&gt; (we can check that the 6 other possibilities lead to &lt;math&gt;(2-1-0)&lt;/math&gt;)<br /> <br /> <br /> In the &lt;math&gt;(2-1-0)&lt;/math&gt; state, we will label the person with &lt;math&gt;\$2&lt;/math&gt; as person A, the person with &lt;math&gt;\$1&lt;/math&gt; as person B, and the person with &lt;math&gt;\$0&lt;/math&gt; as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total &lt;math&gt;2\cdot 2 = 4&lt;/math&gt; ways the money can be redistributed. The only way that the distribution can return to &lt;math&gt;(1-1-1)&lt;/math&gt; is if A gives &lt;math&gt;\$1&lt;/math&gt; to B, and B gives &lt;math&gt;\$1&lt;/math&gt; to C. We check the other possibilities to find that they all lead back to &lt;math&gt;(2-1-0)&lt;/math&gt;. Thus, the probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, while the probability of staying in the &lt;math&gt;(2-1-0)&lt;/math&gt; state is &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> No matter which state we are in, the probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state is always &lt;math&gt;\frac{1}{4}&lt;/math&gt;. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state after the 2019th bell ring. Thus, our answer is simply &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{4}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_22&diff=103812 2019 AMC 10B Problems/Problem 22 2019-02-25T03:46:04Z <p>Vedadehhc: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #22]] and [[2019 AMC 12B Problems|2019 AMC 12B #19]]}}<br /> <br /> ==Problem==<br /> <br /> Raashan, Sylvia, and Ted play the following game. Each starts with &lt;math&gt; \$1&lt;/math&gt;. A bell rings every &lt;math&gt;15&lt;/math&gt; seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives &lt;math&gt;\$1&lt;/math&gt; to that player. What is the probability that after the bell has rung &lt;math&gt;2019&lt;/math&gt; times, each player will have &lt;math&gt;\$1&lt;/math&gt;? (For example, Raashan and Ted may each decide to give &lt;math&gt;\$1&lt;/math&gt; to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have &lt;math&gt;\$0&lt;/math&gt;, Sylvia will have &lt;math&gt;\$2&lt;/math&gt;, and Ted will have &lt;math&gt;\$1&lt;/math&gt;, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their &lt;math&gt; \$1&lt;/math&gt; to, and the holdings will be the same at the end of the second round.)<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{7} \qquad\textbf{(B) } \frac{1}{4} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{2}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> On the first turn, each player starts off with &lt;math&gt;\$1&lt;/math&gt;. Each turn after that, there are only two possibilities: either everyone stays at &lt;math&gt;\$1&lt;/math&gt;, which we will write as &lt;math&gt;(1-1-1)&lt;/math&gt;, or the distribution of money becomes &lt;math&gt;\$2-\$1-\$0&lt;/math&gt; in some order, which we write as &lt;math&gt;(2-1-0)&lt;/math&gt;. We will consider these two states separately.<br /> <br /> In the &lt;math&gt;(1-1-1)&lt;/math&gt; state, each person has two choices for whom to give their dollar to, meaning there are &lt;math&gt;2^3=8&lt;/math&gt; possible ways that the money can be rearranged. Note that there are only two ways that we can reach &lt;math&gt;(1-1-1)&lt;/math&gt; again: <br /> <br /> 1. Raashan gives his money to Sylvia, who gives her money to Ted, who gives his money to Raashan.<br /> <br /> 2. Raashan gives his money to Ted, who gives his money to Sylvia, who gives her money to Raashan.<br /> <br /> Thus, the probability of staying in the &lt;math&gt;(1-1-1)&lt;/math&gt; state is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, while the probability of going to the &lt;math&gt;(2-1-0)&lt;/math&gt; state is &lt;math&gt;\frac{3}{4}&lt;/math&gt; (we can check that the 6 other possibilities lead to &lt;math&gt;(2-1-0)&lt;/math&gt;)<br /> <br /> <br /> In the &lt;math&gt;(2-1-0)&lt;/math&gt; state, we will label the person with &lt;math&gt;\$2&lt;/math&gt; as person A, the person with &lt;math&gt;\$1&lt;/math&gt; as person B, and the person with &lt;math&gt;\$0&lt;/math&gt; as person C. Person A has two options for whom to give money to, and person B has 2 options for whom to give money to, meaning there are total &lt;math&gt;2\cdot 2 = 4&lt;/math&gt; ways the money can be redistributed. The only way that the distribution can return to &lt;math&gt;(1-1-1)&lt;/math&gt; is if A gives &lt;math&gt;\$1&lt;/math&gt; to B, and B gives &lt;math&gt;\$1&lt;/math&gt; to C. We check the other possibilities to find that they all lead back to &lt;math&gt;(2-1-0)&lt;/math&gt;. Thus, the probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state is &lt;math&gt;\frac{1}{4}&lt;/math&gt;, while the probability of staying in the &lt;math&gt;(2-1-0)&lt;/math&gt; state is &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> No matter which state we are in, the probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state is always &lt;math&gt;\frac{1}{4}&lt;/math&gt;. This means that, after the bell rings 2018 times, regardless of what state the money distribution is in, there is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; probability of going to the &lt;math&gt;(1-1-1)&lt;/math&gt; state after the 2019th bell ring. Thus, our answer is simply &lt;math&gt;\boxed{\textbf{(B) } \frac{1}{4}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2019|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Vedadehhc https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_13&diff=103811 2019 AMC 12B Problems/Problem 13 2019-02-25T03:15:10Z <p>Vedadehhc: </p> <hr /> <div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #17]] and [[2019 AMC 12B Problems|2019 AMC 12B #13]]}}<br /> ==Problem==<br /> <br /> A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin &lt;math&gt;k&lt;/math&gt; is &lt;math&gt;2^{-k}&lt;/math&gt; for &lt;math&gt;k = 1,2,3....&lt;/math&gt; What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?&lt;br&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same bin is &lt;math&gt;\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}&lt;/math&gt; (by the geometric series sum formula). Therefore the other two probabilities have to both be &lt;math&gt;\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Suppose the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, for some &lt;math&gt;i \ge 1&lt;/math&gt;. The probability of this occurring is &lt;math&gt;\frac{1}{2^i}&lt;/math&gt;. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is &lt;math&gt;\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}&lt;/math&gt; (by the geometric series sum formula). Thus the probability that the green ball goes in bin &lt;math&gt;i&lt;/math&gt;, and the red ball goes in a bin greater than &lt;math&gt;i&lt;/math&gt;, is &lt;math&gt;\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}&lt;/math&gt;. Summing from &lt;math&gt;i=1&lt;/math&gt; to infinity, we get<br /> <br /> &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;<br /> where we again used the geometric series sum formula. (Alternatively, if this sum equals &lt;math&gt;n&lt;/math&gt;, then by writing out the terms and multiplying both sides by &lt;math&gt;4&lt;/math&gt;, we see &lt;math&gt;4n = n+1&lt;/math&gt;, which gives &lt;math&gt;n = \frac{1}{3}&lt;/math&gt;.)<br /> <br /> ==Solution 3==<br /> The probability that the two balls will go into adjacent bins is &lt;math&gt;\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}&lt;/math&gt; by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of &lt;math&gt;2&lt;/math&gt; from each other is &lt;math&gt;\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}&lt;/math&gt; (again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is &lt;math&gt;\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots&lt;/math&gt;, which, by the geometric series sum formula, is &lt;math&gt;\boxed{\textbf{(C) } \frac{1}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 4 (quick, conceptual)==<br /> Define a win as a ball appearing in higher numbered box.<br /> <br /> Start from the first box. <br /> <br /> There are &lt;math&gt;4&lt;/math&gt; possible results in the box: Red, Green, Red and Green, or none, with an equal probability of &lt;math&gt;\frac{1}{4}&lt;/math&gt; for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if &lt;math&gt;p&lt;/math&gt; is the probability that Red wins, we can write &lt;math&gt;p = \frac{1}{4} + \frac{1}{4}p&lt;/math&gt;: there is a &lt;math&gt;\frac{1}{4}&lt;/math&gt; probability that &quot;Red&quot; wins immediately, a &lt;math&gt;0&lt;/math&gt; probability in the cases &quot;Green&quot; or &quot;Red and Green&quot;, and in the &quot;None&quot; case (which occurs with &lt;math&gt;\frac{1}{4}&lt;/math&gt; probability), we then start again, giving the same probability &lt;math&gt;p&lt;/math&gt;. Hence, solving the equation, we get &lt;math&gt;p = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Write out the infinite geometric series as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, &lt;math&gt;\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots&lt;/math&gt;. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term &lt;math&gt;1&lt;/math&gt;, term &lt;math&gt;3&lt;/math&gt;, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with &lt;math&gt;\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots&lt;/math&gt;. Summing, we get &lt;cmath&gt;\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}&lt;/cmath&gt;<br /> <br /> ==Video Solution==<br /> For those who want a video solution: https://youtu.be/VP7ltu-XEq8<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2019|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2019|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Vedadehhc