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Saudi Arabian Mathematical Competitions, 2015

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle, $\Gamma$ its circumcircle, $I$ its incenter, and $\omega$ a tangent circle to the line $AI$ at $I$ and to the side $BC$ . Prove that the circles $\Gamma$ and $\omega$ are tangent.

Solution

Let $M$ be the midpoint of arc $BC$ not containing $A$ , $E$ the tangent point of $BC$ to $\omega$ , $F$ the second intersection point of $EM$ with $\omega$ . Remember that $M$ is the circumcenter of triangle $BCI$ and therefore $MI = MB$ .

Because $MI$ is tangent to $\omega$ , we have from the power of the point $M$ with respect to $\omega$
$ME \cdot MF = MI^2 = MB^2.$
We deduce that the line $BM$ is tangent to the circumcircle of triangle $BEF$ . Therefore
$\angle BFM = \angle MBE = \angle MAC = \angle BAM.$
This means that point $F$ is on the circle $\Gamma$ .

Because the tangent line to $\Gamma$ at $M$ is parallel to the tangent line to $\omega$ at $E$ and $F, E, M$ are collinear and $F$ is an intersection point of $\Gamma$ and $\omega$ , $F$ is the center of the homothety of the two circles $\omega$ and $\Gamma$ and therefore, they are tangent.

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