Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME Prove it United States

Problem:
Let $ABC$ be a triangle with $AB = 23$ , $BC = 24$ , and $CA = 27$ . Let $D$ be the point on segment $AC$ such that the incircles of triangles $BAD$ and $BCD$ are tangent. Determine the ratio $CD / DA$ .

Solution

Solution:
Answer: $\frac{14}{13}$
Let $X, Z, E$ be the points of tangency of the incircle of $ABD$ to $AB$ , $BD$ , $DA$ respectively. Let $Y, Z, F$ be the points of tangency of the incircle of $CBD$ to $CB$ , $BD$ , $DC$ respectively. We note that
$CB + BD + DC = CY + YB + BZ + ZD + DF + FC = 2(CY) + 2(BY) + 2(DF) = 2(24) + 2(DF)$
by equal tangents, and that similarly
$AB + BD + DA = 2(23) + 2(DE)$
Since $DE = DZ = DF$ by equal tangents, we can subtract the equations above to get that
$CB + CD - AB - DA = 2(24) - 2(23) \Rightarrow CD - DA = 2$
Since we know that $CD + DA = 27$ , we get that $CD = 14$ , $DA = 13$ , so the desired ratio is $\frac{14}{13}$ .

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