2025 SSMO Accuracy Round Problems/Problem 10

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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2025 SSMO Accuracy Round Problems/Problem 10

Problem

Solution