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Difference between revisions of "2024 SSMO Team Round Problems/Problem 4"

(Created page with "==Problem== Let <math>ABC</math> be a right triangle with circumcenter <math>O</math> and incenter <math>I</math> such that <math>\angle ABC = 90^{\circ}</math> and <math>\fr...")
 
 
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==Solution==
 
==Solution==
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We will use analytic geometry. WLOG, let <math>B = (0,0),A = (0,6),C = (8,0).</math> without loss of generality due to rigid transformations or dilations. Since the area of <math>ABC</math> is <math>\frac{1}{2}\cdot8\cdot6 = 24,</math> the inradius is <math>\frac{2\cdot24}{6+8+10} = 2,</math> meaning <math>I = (2,2).</math> Since <math>ABC</math> is a right triangle, the <math>O</math> must be the midpoint of <math>AC.</math> So, <math>D</math> and <math>E</math> are the midpoints of <math>AB</math> and <math>BC,</math> respectively. Now, as <math>ADO</math> and <math>OEC</math> are both similar to <math>ABC,</math> we can easily compute <math>\omega_1 = (1,4)</math> and <math>\omega_2 = (5,1).</math> From the shoelace theorem, the area of <math>\omega_1\omega_2I</math> is <math>\frac{5}{2},</math> meaning our answer is <cmath>\frac{\frac{5}{2}}{24} = \frac{5}{48}\implies 5+48 = \boxed{53}.</cmath>
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~SMO_Team

Latest revision as of 14:36, 10 September 2025

Problem

Let $ABC$ be a right triangle with circumcenter $O$ and incenter $I$ such that $\angle ABC = 90^{\circ}$ and $\frac{AB}{BC} = \frac{3}{4}$. Let $D$ the projection of $O$ onto $AB$, and let $E$ be the projection of $O$ onto $BC$. Denote $\omega_{1}$ be the incenter of $ADO$ and $\omega_{2}$ as the incenter of $OEC$. If $\frac{[\omega_{1}\omega_{2}I]}{[ABC]}=\frac{m}{n},$ for relatively prime positive integers $m$ and $n,$ find $m+n.$

Solution

We will use analytic geometry. WLOG, let $B = (0,0),A = (0,6),C = (8,0).$ without loss of generality due to rigid transformations or dilations. Since the area of $ABC$ is $\frac{1}{2}\cdot8\cdot6 = 24,$ the inradius is $\frac{2\cdot24}{6+8+10} = 2,$ meaning $I = (2,2).$ Since $ABC$ is a right triangle, the $O$ must be the midpoint of $AC.$ So, $D$ and $E$ are the midpoints of $AB$ and $BC,$ respectively. Now, as $ADO$ and $OEC$ are both similar to $ABC,$ we can easily compute $\omega_1 = (1,4)$ and $\omega_2 = (5,1).$ From the shoelace theorem, the area of $\omega_1\omega_2I$ is $\frac{5}{2},$ meaning our answer is \[\frac{\frac{5}{2}}{24} = \frac{5}{48}\implies 5+48 = \boxed{53}.\]

~SMO_Team

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