2024 AIME I Problems/Problem 10

Problem

Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $P$ , and let $\overline{AP}$ intersect $\omega$ at $D$ . Find $AD$ , if $AB=5$ , $BC=9$ , and $AC=10$ .

Solution 1

From the tangency condition we have $\let\angle BCD = \let\angle CBD = \let\angle A$ . With LoC we have $\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}$ and $\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}$ . Then, $CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}$ . Using LoC we can find $AD$ : $AD^2 = AC^2 + CD^2 - 2(AC)(CD)\cos(A+C) = 10^2+(\frac{225}{22})^2 + 2(10)\frac{225}{22}\cos(B) = 100 + \frac{225^2}{22^2} + 2(10)\frac{225}{22}*\frac{1}{15} = \frac{5^4*13^2}{484}$ . Thus, $AD = \frac{5^2*13}{22}$ . By Power of a Point, $DP*AD = CD^2$ so $DP*\frac{5^2*13}{22} = (\frac{225}{22})^2$ which gives $DP = \frac{5^2*9^2}{13*22}$ . Finally, we have $AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}$ .

~angie.

Solution 2

Well know $AP$ is the symmedian, which implies $\triangle{ABP}\sim \triangle{AMC}$ where $M$ is the midpoint of $BC$ . By Appolonius theorem, $AM=\frac{13}{2}$ . Thus, we have $\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}$

~Bluesoul

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2024 AIME I Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See also