https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Daniyalqazi2&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-18T13:29:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_9&diff=86944 2013 AIME I Problems/Problem 9 2017-08-08T04:51:05Z <p>Daniyalqazi2: /* Solution 1 */</p> <hr /> <div>==Problem 9==<br /> A paper equilateral triangle &lt;math&gt;ABC&lt;/math&gt; has side length &lt;math&gt;12&lt;/math&gt;. The paper triangle is folded so that vertex &lt;math&gt;A&lt;/math&gt; touches a point on side &lt;math&gt;\overline{BC}&lt;/math&gt; a distance &lt;math&gt;9&lt;/math&gt; from point &lt;math&gt;B&lt;/math&gt;. The length of the line segment along which the triangle is folded can be written as &lt;math&gt;\frac{m\sqrt{p}}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import cse5;<br /> size(12cm);<br /> pen tpen = defaultpen + 1.337;<br /> real a = 39/5.0;<br /> real b = 39/7.0;<br /> pair B = MP(&quot;B&quot;, (0,0), dir(200));<br /> pair A = MP(&quot;A&quot;, (9,0), dir(-80));<br /> pair C = MP(&quot;C&quot;, (12,0), dir(-20));<br /> pair K = (6,10.392);<br /> pair M = (a*B+(12-a)*K) / 12;<br /> pair N = (b*C+(12-b)*K) / 12;<br /> draw(B--M--N--C--cycle, tpen);<br /> draw(M--A--N--cycle);<br /> fill(M--A--N--cycle, mediumgrey);<br /> pair shift = (-20.13, 0);<br /> pair B1 = MP(&quot;B&quot;, B+shift, dir(200));<br /> pair A1 = MP(&quot;A&quot;, K+shift, dir(90));<br /> pair C1 = MP(&quot;C&quot;, C+shift, dir(-20));<br /> draw(A1--B1--C1--cycle, tpen);&lt;/asy&gt;<br /> <br /> == Solution 1 ==<br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, where the paper is folded.<br /> <br /> Let &lt;math&gt;D&lt;/math&gt; be the point on &lt;math&gt;\overline{BC}&lt;/math&gt; where the folded &lt;math&gt;A&lt;/math&gt; touches it.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;x&lt;/math&gt; be the lengths &lt;math&gt;AP&lt;/math&gt;, &lt;math&gt;AQ&lt;/math&gt;, and &lt;math&gt;PQ&lt;/math&gt;, respectively.<br /> <br /> We have &lt;math&gt;PD = a&lt;/math&gt;, &lt;math&gt;QD = b&lt;/math&gt;, &lt;math&gt;BP = 12 - a&lt;/math&gt;, &lt;math&gt;CQ = 12 - b&lt;/math&gt;, &lt;math&gt;BD = 9&lt;/math&gt;, and &lt;math&gt;CD = 3&lt;/math&gt;.<br /> <br /> Using the Law of Cosines on &lt;math&gt;BPD&lt;/math&gt;:<br /> <br /> &lt;math&gt;a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a&lt;/math&gt;<br /> <br /> &lt;math&gt;a = \frac{39}{5}&lt;/math&gt;<br /> <br /> Using the Law of Cosines on &lt;math&gt;CQD&lt;/math&gt;:<br /> <br /> &lt;math&gt;b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b&lt;/math&gt;<br /> <br /> &lt;math&gt;b = \frac{39}{7}&lt;/math&gt;<br /> <br /> Using the Law of Cosines on &lt;math&gt;DPQ&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^{2} = a^{2} + b^{2} - 2ab \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{39 \sqrt{39}}{35}&lt;/math&gt;<br /> <br /> The solution is &lt;math&gt;39 + 39 + 35 = \boxed{113}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Proceed with the same labeling as in Solution 1. <br /> <br /> &lt;math&gt;\angle B = \angle C = \angle A = \angle PDQ = 60^\circ&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;\angle PDB = \angle CQD&lt;/math&gt;.<br /> <br /> Similarly, &lt;math&gt;\angle BPD = \angle QDC&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;\bigtriangleup BPD&lt;/math&gt; and &lt;math&gt;\bigtriangleup CDQ&lt;/math&gt; are similar triangles, so<br /> <br /> &lt;math&gt;\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}&lt;/math&gt;.<br /> <br /> Solving this system of equations yields &lt;math&gt;a = \frac{39}{5}&lt;/math&gt; and &lt;math&gt;b = \frac{39}{7}&lt;/math&gt;.<br /> <br /> Using the Law of Cosines on &lt;math&gt;APQ&lt;/math&gt;:<br /> <br /> &lt;math&gt;x^{2} = a^{2} + b^{2} - 2ab \cos{60}&lt;/math&gt;<br /> <br /> &lt;math&gt;x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})&lt;/math&gt;<br /> <br /> &lt;math&gt;x = \frac{39 \sqrt{39}}{35}&lt;/math&gt;<br /> <br /> The solution is &lt;math&gt;39 + 39 + 35 = \boxed{113}&lt;/math&gt;.<br /> <br /> == Solution 3 (Coordinate Bash) ==<br /> <br /> e let the original position of &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, and the position of &lt;math&gt;A&lt;/math&gt; after folding be &lt;math&gt;D&lt;/math&gt;. Also, we put the triangle on the coordinate plane such that &lt;math&gt;A=(0,0)&lt;/math&gt;, &lt;math&gt;B=(-6,-6\sqrt3)&lt;/math&gt;, &lt;math&gt;C=(6,-6\sqrt3)&lt;/math&gt;, and &lt;math&gt;D=(3,-6\sqrt3)&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> pen tpen = defaultpen + 1.337;<br /> real a = 39/5.0;<br /> real b = 39/7.0;<br /> pair B = MP(&quot;B&quot;, (0,0), dir(200));<br /> pair A = (9,0);<br /> pair C = MP(&quot;C&quot;, (12,0), dir(-20));<br /> pair K = (6,10.392);<br /> pair M = (a*B+(12-a)*K) / 12;<br /> pair N = (b*C+(12-b)*K) / 12;<br /> draw(B--M--N--C--cycle);<br /> draw(M--A--N--cycle);<br /> label(&quot;$D$&quot;, A, S);<br /> pair X = (6,6*sqrt(3));<br /> draw(B--X--C);<br /> label(&quot;$A$&quot;,X,dir(90));<br /> draw(A--X);<br /> &lt;/asy&gt;<br /> <br /> Note that since &lt;math&gt;A&lt;/math&gt; is reflected over the fold line to &lt;math&gt;D&lt;/math&gt;, the fold line is the perpendicular bisector of &lt;math&gt;AD&lt;/math&gt;. We know &lt;math&gt;A=(0,0)&lt;/math&gt; and &lt;math&gt;D=(3,-6\sqrt3)&lt;/math&gt;. The midpoint of &lt;math&gt;AD&lt;/math&gt; (which is a point on the fold line) is &lt;math&gt;(\tfrac32, -3\sqrt3)&lt;/math&gt;. Also, the slope of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;\frac{-6\sqrt3}{3}=-2\sqrt3&lt;/math&gt;, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of &lt;math&gt;AD&lt;/math&gt;, or &lt;math&gt;\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}&lt;/math&gt;. Then, using point slope form, the equation of the fold line is<br /> &lt;cmath&gt;y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)&lt;/cmath&gt;&lt;cmath&gt;y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;<br /> Note that the equations of lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are &lt;math&gt;y=\sqrt3x&lt;/math&gt; and &lt;math&gt;y=-\sqrt3x&lt;/math&gt;, respectively. We will first find the intersection of &lt;math&gt;AB&lt;/math&gt; and the fold line by substituting for &lt;math&gt;y&lt;/math&gt;:<br /> &lt;cmath&gt;\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;&lt;cmath&gt;\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}&lt;/cmath&gt;<br /> Therefore, the point of intersection is &lt;math&gt;\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)&lt;/math&gt;. Now, lets find the intersection with &lt;math&gt;AC&lt;/math&gt;. Substituting for &lt;math&gt;y&lt;/math&gt; yields<br /> &lt;cmath&gt;-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}&lt;/cmath&gt;&lt;cmath&gt;\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}&lt;/cmath&gt;<br /> Therefore, the point of intersection is &lt;math&gt;\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)&lt;/math&gt;. Now, we just need to use the distance formula to find the distance between &lt;math&gt;\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)&lt;/math&gt; and &lt;math&gt;\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)&lt;/math&gt;.<br /> &lt;cmath&gt;\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}&lt;/cmath&gt;<br /> The number 39 is in all of the terms, so let's factor it out:<br /> &lt;cmath&gt;39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}&lt;/cmath&gt;&lt;cmath&gt;\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}&lt;/cmath&gt;<br /> Therefore, our answer is &lt;math&gt;39+39+35=\boxed{113}&lt;/math&gt;, and we are done.<br /> <br /> Solution by nosaj.<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Daniyalqazi2 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_19&diff=79887 2013 AMC 8 Problems/Problem 19 2016-08-07T16:23:01Z <p>Daniyalqazi2: Hannah had two a's</p> <hr /> <div>==Problem==<br /> Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Hannah, Cassie, Bridget} \qquad \textbf{(B)}\ \text{Hannah, Bridget, Cassie} \\ \qquad \textbf{(C)}\ \text{Cassie, Bridget, Hannah} \qquad \textbf{(D)}\ \text{Cassie, Hannah, Bridget} \\ \qquad \textbf{(E)}\ \text{Bridget, Cassie, Hannah}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, if Hannah did worse than Bridget, there is no way Bridget could have known that she didn't get the highest in the class. Therefore, Hannah did better than Bridget, so our order is &lt;math&gt;\boxed{\textbf{(D)}\ \text{Cassie, Hannah, Bridget}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Daniyalqazi2