HMMT February, 2020

Geometry Difficulty 5.9 AIME, harder Prove it United States

Problem:
Let $\triangle ABC$ be a triangle inscribed in a unit circle with center $O$ . Let $I$ be the incenter of $\triangle ABC$ , and let $D$ be the intersection of $BC$ and the angle bisector of $\angle BAC$ . Suppose that the circumcircle of $\triangle ADO$ intersects $BC$ again at a point $E$ such that $E$ lies on $IO$ . If $\cos A=\frac{12}{13}$ , find the area of $\triangle ABC$ .

Solution

Solution:
Consider the following lemma:
Lemma. $AD \perp EO$ .
Proof. By the Shooting Lemma, the reflection of the midpoint $M$ of arc $BC$ not containing $A$ over $BC$ lies on $(ADO)$ . Hence
$\measuredangle ADE + \measuredangle DEO = \measuredangle MDC + \measuredangle DM' O = \measuredangle MDC + \measuredangle M' MD = 90^\circ$
This is enough to imply $AD \perp EO$ .

Thus $I$ is the foot from $O$ onto $AD$ . Now
$AI^2 + IO^2 = AO^2$
By Euler's formula,
$\left(\frac{r}{\sin \frac{A}{2}}\right)^2 + R^2 - 2Rr = R^2$
Hence
$r = 2R \sin^2 \frac{A}{2}$
Then
$s = a + \frac{r}{\tan \frac{A}{2}} = a + R \sin A = 3R \sin A$
and
$[ABC] = r s = \left(2R \sin^2 \frac{A}{2}\right)(3R \sin A)$
Since $R=1$ , we get
$[ABC] = 3(1-\cos A) \sin A$
Plugging in $\sin A = \frac{5}{13}$ and $\cos A = \frac{12}{13}$ , we get
$[ABC] = 3 \cdot \frac{1}{13} \cdot \frac{5}{13} = \frac{15}{169}$

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