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2024 AIME I Problems/Problem 10

Revision as of 13:25, 3 February 2024 by Mathwiz 1207 (talk | contribs)

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

Solution 3

Extend sides $\overline{AB}$ and $\overline{AC}$ to points $E$ and $F$, respectively, such that $B$ and $C$ are the feet of the altitudes in $\triangle AEF$. Denote the feet of the altitude from $A$ to $\overline{EF}$ as $X$, and let $H$ denote the orthocenter of $\triangle AEF$. Call $M$ the midpoint of segment $\overline{EF}$. By the Three Tangents Lemma, we have that $MB$ and $MC$ are both tangents to $(ABC)$ $\implies$ $M = P$, and since $M$ is the midpoint of $\overline{EF}$, $MF = MB$. Additionally, by angle chasing, we get that: \[\angle ABC \cong \angle AHC \cong \angle EHX\] Also, \[\angle EHX = 90 ^\circ - \angle HEF = 90 ^\circ - (90 ^\circ - \angle AFE) = \angle AFE\] Furthermore, \[AB = AF \cdot \cos(A)\] From this, we see that $\triangle ABC \sim \triangle AFE$ with a scale factor of $\cos(A)$. By the Law of Cosines, \[\cos(A) = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}\] Thus, we can find that the side lengths of $\triangle AEF$ are $\frac{250}{11}, \frac{125}{11}, \frac{225}{11}$. Then, by Stewart's theorem, $AM = \frac{13 \cdot 25}{22}$. By Power of a Point, \[\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}\] \[\frac{225}{22} \cdot \frac{225}{22} = \overline{MD} \cdot \frac{13 \cdot 25}{22} \implies \overline{MD} = \frac{225 \cdot 9}{22 \cdot 13}\] Thus, \[AD = AM - MD = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}\] Therefore, the answer is $\boxed{113}$.

~mathwiz_1207

See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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