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

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

Revision as of 02:44, 11 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

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