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Difference between revisions of "2024 AIME II Problems/Problem 10"

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==Problem==
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Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math> with <math>\overline{IA}\perp\overline{OI}</math>, circumradius <math>13</math>, and inradius <math>6</math>. Find <math>AB\cdot AC</math>.
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==Solution==
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By Euler's formula <math>OI^{2}=R(R-2r)</math>, we have <math>OI^{2}=13(13-12)=13</math>. Thus, by the Pythagorean theorem, <math>AI^{2}=13^{2}-13=156</math>. Let <math>AI\cap(ABC)=M</math>; notice <math>\triangle AOM</math> is isosceles and <math>\overline{OI}\perp\overline{AM}</math> which is enough to imply that <math>I</math> is the midpoint of <math>\overline{AM}</math>, and <math>M</math> itself is the midpoint of <math>II_{a}</math> where <math>I_{a}</math> is the <math>A</math>-excenter of <math>\triangle ABC</math>. Therefore, <math>AI=IM=MI_{a}=\sqrt{156}</math> and <cmath>AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.</cmath>
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Note that this problem is extremely similar to [[2019 CIME I Problems/Problem 14|2019 CIME I/14]].
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==See also==
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{{AIME box|year=2024|num-b=9|num-a=11|n=II}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Revision as of 16:09, 8 February 2024

Problem

Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$, circumradius $13$, and inradius $6$. Find $AB\cdot AC$.

Solution

By Euler's formula $OI^{2}=R(R-2r)$, we have $OI^{2}=13(13-12)=13$. Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$. Let $AI\cap(ABC)=M$; notice $\triangle AOM$ is isosceles and $\overline{OI}\perp\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\overline{AM}$, and $M$ itself is the midpoint of $II_{a}$ where $I_{a}$ is the $A$-excenter of $\triangle ABC$. Therefore, $AI=IM=MI_{a}=\sqrt{156}$ and \[AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.\]

Note that this problem is extremely similar to 2019 CIME I/14.

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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