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Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 10"

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==Solution==
 
==Solution==
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<asy>
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import geometry;
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unitsize(2cm);
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point A=(0,0);
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point B=dir(80);
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point C=2*dir(20);
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point D=4*dir(-40);
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point E=8*dir(-100);
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draw(A--B--C--cycle);
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draw(A--D--C);
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draw(A--E--D);
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dot(A); dot(B); dot(C);
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</asy>
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The main claim is that <math>P</math> is the the second intersection point of <math>(ABC)</math> and <math>(ADE)</math>.

Latest revision as of 23:56, 20 September 2025

Problem

Let $ABCDE$ be a convex pentagon with $\angle{BAC} = \angle{CAD} = \angle{DAE}$ and $\angle{ABC} = \angle{ACD} = \angle{ADE}$. Let $BD$ and $CE$ meet at $P$. Given that $BC = 6$, $\sin{\angle{BAC}} = \tfrac{3}{5}$, and $\tfrac{AC}{AB} = 5$, the length of $AP$ can be expressed as $\tfrac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers such that $n$ is square-free. Find $m+n$.

Solution

[asy] import geometry; unitsize(2cm); point A=(0,0); point B=dir(80); point C=2*dir(20); point D=4*dir(-40); point E=8*dir(-100); draw(A--B--C--cycle); draw(A--D--C); draw(A--E--D); dot(A); dot(B); dot(C); [/asy]

The main claim is that $P$ is the the second intersection point of $(ABC)$ and $(ADE)$.

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