Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

Triangle $A B C$ has $A B=4, B C=3$ , and a right angle at $B$ . Circles $\omega_{1}$ and $\omega_{2}$ of equal radii are drawn such that $\omega_{1}$ is tangent to $A B$ and $A C, \omega_{2}$ is tangent to $B C$ and $A C$ , and $\omega_{1}$ is tangent to $\omega_{2}$ . Find the radius of $\omega_{1}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Denote by $r$ the common radius of $\omega_{1}, \omega_{2}$ , and let $O_{1}, O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ respectively. Suppose $\omega_{i}$ hits $A C$ at $B_{i}$ for $i=1,2$ , so that $O_{1} O_{2}=B_{1} B_{2}=2 r$ . Extend angle bisector $A O_{1}$ to hit $B C$ at $P$ . By the angle bisector theorem and triangle similarity $\triangle A B_{1} O_{1} \sim \triangle A B P$ , we deduce $\frac{r}{A B_{1}}=\frac{B P}{A B}=\frac{3}{4+5}$ . Similarly, $\frac{r}{C B_{2}}=\frac{4}{3+5}$ , so $5=A C=A B_{1}+B_{1} B_{2}+B_{2} C=3 r+2 r+2 r=7 r$ or $r=\frac{5}{7}$ .

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