Competition problem — Geometry — Maths Olympiad Prep

Library / /650 of 860

Geometry Difficulty 5.3 AIME, harder Find the answer

In right triangle $A B C$ , a point $D$ is on hypotenuse $A C$ such that $B D \perp A C$ . Let $\omega$ be a circle with center $O$ , passing through $C$ and $D$ and tangent to line $A B$ at a point other than $B$ . Point $X$ is chosen on $B C$ such that $A X \perp B O$ . If $A B=2$ and $B C=5$ , then $B X$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100 a+b$ .

Your answer

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

Solution

Note that since $A D \cdot A C=A B^{2}$ , we have the tangency point of $\omega$ and $A B$ is $B^{\prime}$ , the reflection of $B$ across $A$ . Let $Y$ be the second intersection of $\omega$ and $B C$ . Note that by power of point, we have $B Y \cdot B C=B B^{\prime 2}=4 A B^{2} \Longrightarrow B Y=\frac{4 A B^{2}}{B C}$ . Note that $A X$ is the radical axis of $\omega$ and the degenerate circle at $B$ , so we have $X B^{2}=X Y \cdot X C$ , so $B X^{2}=(B C-B X)(B Y-B X)=B X^{2}-B X(B C+B Y)+B C \cdot B Y$ This gives us $B X=\frac{B C \cdot B Y}{B C+B Y}=\frac{4 A B^{2} \cdot B C}{4 A B^{2}+B C^{2}}=\frac{80}{41}$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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