Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Triangle $ABC$ has incircle $\omega$ which touches $AB$ at $C_1$ , $BC$ at $A_1$ , and $CA$ at $B_1$ . Let $A_2$ be the reflection of $A_1$ over the midpoint of $BC$ , and define $B_2$ and $C_2$ similarly. Let $A_3$ be the intersection of $AA_2$ with $\omega$ that is closer to $A$ , and define $B_3$ and $C_3$ similarly. If $AB=9$ , $BC=10$ , and $CA=13$ , find $\left[A_3 B_3 C_3\right] / [ABC]$ . (Here $[XYZ]$ denotes the area of triangle $XYZ$ .)

Solution

Solution:

Notice that $A_2$ is the point of tangency of the excircle opposite $A$ to $BC$ . Therefore, by considering the homothety centered at $A$ taking the excircle to the incircle, we notice that $A_3$ is the intersection of $\omega$ and the tangent line parallel to $BC$ . It follows that $A_1 B_1 C_1$ is congruent to $A_3 B_3 C_3$ by reflecting through the center of $\omega$ . We therefore need only find $\left[A_1 B_1 C_1\right] / [ABC]$ . Since
$\frac{\left[A_1 BC_1\right]}{[ABC]} = \frac{A_1 B \cdot BC_1}{AB \cdot BC} = \frac{((9+10-13)/2)^2}{9 \cdot 10} = \frac{1}{10}$
and likewise $\left[A_1 B_1 C\right] / [ABC] = 49/130$ and $\left[AB_1 C_1\right] / [ABC] = 4/13$ , we get that

$\text{[A 3}$ B_3 ${\left. C_3\right]}{[ABC]}$ = 1 - $\frac{1}{10} - \frac{49}{130} - \frac{4}{13} = \frac{14}{65}.$

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