HMMT February

Consider the following lemma: Lemma. $A D \perp E O$. Proof. By the Shooting Lemma, the reflection of the midpoint $M$ of arc $B C$ not containing $A$ over $B C$ lies on $(A D O)$. Hence $\measuredangle A D E+\measuredangle D E O=\measuredangle M D C+\measuredangle D M^{\prime} O=\measuredangle M D C+\measuredangle M^{\prime} M D=90^{\circ}$. This is enough to imply $A D \perp E O$. Thus $I$ is the foot from $O$ onto $A D$. Now $A I^{2}+I O^{2}=A O^{2}$. By Euler's formula, $\left(\frac{r}{\sin \frac{A}{2}}\right)^{2}+R^{2}-2 R r=R^{2}$. Hence $r=2 R \sin ^{2} \frac{A}{2}$. Then $s=a+\frac{r}{\tan \frac{A}{2}}=a+R \sin A=3 R \sin A$ and $[A B C]=r s=\left(2 R \sin ^{2} \frac{A}{2}\right)(3 R \sin A)$. Since $R=1$, we get $[A B C]=3(1-\cos A) \sin A$. Plugging in $\sin A=\frac{5}{13}$ and $\cos A=\frac{12}{13}$, we get $[A B C]=3 \cdot \frac{1}{13} \cdot \frac{5}{13}=\frac{15}{169}$.