Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Find the answer

Let $A B C$ be an acute triangle with circumcenter $O$ such that $A B=4, A C=5$ , and $B C=6$ . Let $D$ be the foot of the altitude from $A$ to $B C$ , and $E$ be the intersection of $A O$ with $B C$ . Suppose that $X$ is on $B C$ between $D$ and $E$ such that there is a point $Y$ on $A D$ satisfying $X Y \parallel A O$ and $Y O \perp A X$ . Determine the length of $B X$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $A X$ intersect the circumcircle of $\triangle A B C$ again at $K$ . Let $O Y$ intersect $A K$ and $B C$ at $T$ and $L$ , respectively. We have $\angle L O A=\angle O Y X=\angle T D X=\angle L A K$ , so $A L$ is tangent to the circumcircle. Furthermore, $O L \perp A K$ , so $\triangle A L K$ is isosceles with $A L=A K$ , so $A K$ is also tangent to the circumcircle. Since $B C$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L, C L$ is a symmedian of $\triangle A C K$ . Then $A K$ is a symmedian of $\triangle A B C$ . Then we can use $\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}$ to compute $B X=\frac{96}{41}$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.