HMMT November

Geometry Difficulty 4.9 AIME Find the answer

Let $A B C$ be a triangle with $A B=5, B C=8, C A=11$ . The incircle $\omega$ and $A$ -excircle $^{1} \Gamma$ are centered at $I_{1}$ and $I_{2}$ , respectively, and are tangent to $B C$ at $D_{1}$ and $D_{2}$ , respectively. Find the ratio of the area of $\triangle A I_{1} D_{1}$ to the area of $\triangle A I_{2} D_{2}$ .

Your answer

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

Solution

Let $D_{1}^{\prime}$ and $D_{2}^{\prime}$ be the points diametrically opposite $D_{1}$ and $D_{2}$ on the incircle and $A$ -excircle, respectively. As $I_{x}$ is the midpoint of $D_{x}$ and $D_{x}^{\prime}$ , we have $\frac{\left[A I_{1} D_{1}\right]}{\left[A I_{2} D_{2}\right]}=\frac{\left[A D_{1} D_{1}^{\prime}\right]}{\left[A D_{2} D_{2}^{\prime}\right]}$ Now, $\triangle A D_{1} D_{1}^{\prime}$ and $\triangle A D_{2} D_{2}^{\prime}$ are homothetic with ratio $\frac{r}{r_{A}}=\frac{s-a}{s}$ , where $r$ is the inradius, $r_{A}$ is the $A$ -exradius, and $s$ is the semiperimeter. Our answer is thus $\left(\frac{s-a}{s}\right)^{2}=\left(\frac{4}{12}\right)=\frac{1}{9}$

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