2024 AIME II Problems/Problem 10

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.

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

Page

Toolbox

Search

2024 AIME II Problems/Problem 10

Problem

Solution

See also