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, 2019

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $ABC$ be a triangle with $AB = 5$ , $BC = 8$ , $CA = 11$ . The incircle $\omega$ and $A$ -excircle $\Gamma$ are centered at $I_1$ and $I_2$ , respectively, and are tangent to $BC$ 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

Solution:
Let $D_1'$ and $D_2'$ 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 $D_x'$ , we have
$\frac{[A I_1 D_1]}{[A I_2 D_2]} = \frac{[A D_1 D_1']}{[A D_2 D_2']}$
Now, $\triangle A D_1 D_1'$ and $\triangle A D_2 D_2'$ 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
$\left(\frac{s-a}{s}\right)^2 = \left(\frac{4}{12}\right)^2 = \frac{1}{9}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.