2024 SSMO Team Round Problems/Problem 5

Problem

Let $ABC$ be a triangle with $AB=AC=5$ and $BC=6$ . Let $\omega_1$ be the circumcircle of $ABC$ and let $\omega_2$ be the circle externally tangent to $\omega_1$ and tangent to rays $AB$ and $AC$ . If the distance between the centers of $\omega_1$ and $\omega_2$ can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m+n$ .

Solution

Let $D$ be the midpoint of $BC$ and let $E$ be the point at which $\omega_1$ and $\omega_2$ are tangent. By symmetry, $O_1$ and $O_2$ , the centers of $\omega_1$ and $\omega_2$ , are on line $ADE$ . Now, $\triangle ADC \sim \triangle ACE$ , so the radius of $\omega_1$ is $r_1=\frac{5\cdot\frac54}2=\frac{25}{8}.$ If $F$ is the point at which $\omega_2$ and $AC$ are tangent, then $AO_2F$ is a 3-4-5 right triangle. If $r_2$ is the radius of $\omega_2$ , we find that $\frac{r_2}{r_2+\frac{25}4}=\frac35$ so $r_2=\frac{75}{8}$ . Therefore, the final answer is $r_1+r_2=\frac{25}{2}\implies 25+2=\boxed{27}.$

~SMO_Team

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2024 SSMO Team Round Problems/Problem 5

Problem

Solution