2024 AIME I Problems/Problem 10: Difference between revisions

Revision as of 13:56, 3 February 2024

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

Video Solution 1 by OmegaLearn.org

https://youtu.be/heryP002bp8

@@ Line 31: / Line 31: @@
 ~mathwiz_1207
+==Video Solution 1 by OmegaLearn.org==
+https://youtu.be/heryP002bp8
 ==See also==

2024 AIME I (Problems • Answer Key • Resources)
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