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Difference between revisions of "2024 SSMO Team Round Problems/Problem 5"
 
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==Solution==
 
==Solution==
 
Let <math>D</math> be the midpoint of <math>BC</math> and let <math>E</math> be the point at which <math>\omega_1</math> and <math>\omega_2</math> are tangent. By symmetry, <math>O_1</math> and <math>O_2</math>, the centers of <math>\omega_1</math> and <math>\omega_2</math>, are on line <math>ADE</math>. Now, <math>\triangle ADC \sim \triangle ACE</math>, so the radius of <math>\omega_1</math> is <cmath>r_1=\frac{5\cdot\frac54}2=\frac{25}{8}.</cmath> If <math>F</math> is the point at which <math>\omega_2</math> and <math>AC</math> are tangent, then <math>AO_2F</math> is a 3-4-5 right triangle. If <math>r_2</math> is the radius of <math>\omega_2</math>, we find that <cmath>\frac{r_2}{r_2+\frac{25}4}=\frac35</cmath> so <math>r_2=\frac{75}{8}</math>. Therefore, the final answer is <cmath>r_1+r_2=\frac{25}{2}\implies 25+2=\boxed{27}.</cmath>
 
Let <math>D</math> be the midpoint of <math>BC</math> and let <math>E</math> be the point at which <math>\omega_1</math> and <math>\omega_2</math> are tangent. By symmetry, <math>O_1</math> and <math>O_2</math>, the centers of <math>\omega_1</math> and <math>\omega_2</math>, are on line <math>ADE</math>. Now, <math>\triangle ADC \sim \triangle ACE</math>, so the radius of <math>\omega_1</math> is <cmath>r_1=\frac{5\cdot\frac54}2=\frac{25}{8}.</cmath> If <math>F</math> is the point at which <math>\omega_2</math> and <math>AC</math> are tangent, then <math>AO_2F</math> is a 3-4-5 right triangle. If <math>r_2</math> is the radius of <math>\omega_2</math>, we find that <cmath>\frac{r_2}{r_2+\frac{25}4}=\frac35</cmath> so <math>r_2=\frac{75}{8}</math>. Therefore, the final answer is <cmath>r_1+r_2=\frac{25}{2}\implies 25+2=\boxed{27}.</cmath>
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~SMO_Team

Latest revision as of 14:37, 10 September 2025

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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