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Difference between revisions of "2024 AIME II Problems/Problem 10"
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Note that this problem is extremely similar to [[2019 CIME I Problems/Problem 14|2019 CIME I/14]].
 
Note that this problem is extremely similar to [[2019 CIME I Problems/Problem 14|2019 CIME I/14]].
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==Solution 2==
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Denote <math>AB=a, AC=b, BC=c</math>. By the given condition, <math>\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6</math>, where <math>A</math> is the area of <math>\triangle{ABC}</math>.
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Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By Fact 5, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>.
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Thus, we have <math>\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c</math>. Now, we have <math>\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}</math>
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~Bluesoul
  
 
==See also==
 
==See also==

Revision as of 17:58, 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.


Solution 2

Denote $AB=a, AC=b, BC=c$. By the given condition, $\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6$, where $A$ is the area of $\triangle{ABC}$.

Moreover, since $OI\bot AI$, the second intersection of the line $AI$ and $(ABC)$ is the reflection of $A$ about $I$, denote that as $D$. By Fact 5, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$.

Thus, we have $\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c$. Now, we have $\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}$

~Bluesoul

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

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