HMMT-Nov 2019 guts, Problem 19

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 $\triangle A I_{2} D_{2}$.

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 \[ \left(\frac{s-a}{s}\right)^{2}=\left(\frac{4}{12}\right)=\frac{1}{9} \] $\fbox{\frac{1}{9}}$.

Answer: \frac{1}{9}