HMMT November

Geometry Difficulty 4.7 AIME Find the answer

In triangle $A B C$ with $A B=8$ and $A C=10$ , the incenter $I$ is reflected across side $A B$ to point $X$ and across side $A C$ to point $Y$ . Given that segment $X Y$ bisects $A I$ , compute $B C^{2}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let $E, F$ be the tangency points of the incircle to sides $A C, A B$ , respectively. Due to symmetry around line $A I, A X I Y$ is a rhombus. Therefore $\angle X A I=2 \angle E A I=2\left(90^{\circ}-\angle E I A\right)=180^{\circ}-2 \angle X A I$ which implies that $60^{\circ}=\angle X A I=2 \angle E A I=\angle B A C$ . By the law of cosines, $B C^{2}=8^{2}+10^{2}-2 \cdot 8 \cdot 10 \cdot \cos 60^{\circ}=84$

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