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Geometry Difficulty 5.2 AIME, harder Prove it India

Let $ABC$ be a triangle with $AB = AC$ , and let $\Gamma$ be its circumcircle. Suppose the incircle $\gamma$ of $ABC$ moves (slides) on $BC$ in the direction of $B$ . Prove that when $\gamma$ touches $\Gamma$ internally, it also touches the altitude through $A$ .

Solution

Let $\gamma'$ be the position of $\gamma$ , when it touches $\Gamma$ internally, and let $K$ be its centre. Let $O$ be the circumcentre and $I$ be the in-centre of $ABC$ . Since $AB = AC$ , both of these lie on the altitude $AD$ . If $T$ is the point of contact of $\Gamma$ and $\gamma'$ , then $T, K, O$ are collinear. Hence $OK = OT - KT = R - r$ , where $R$ and $r$ are respectively the circumradius and inradius of $ABC$ . Note that $K$ and $I$ are at same distance $r$ from $BC$ . Thus $KI$ is perpendicular to $AD$ at $I$ . Using the right-angled triangle $OKI$ , we have $OK^2 = OI^2 + IK^2$ . But $OI^2 = R^2 - 2Rr$ . Hence we obtain

$IK^2 = OK^2 - OI^2 = (R-r)^2 - (R^2 - 2Rr) = r^2.$

Thus $IK = r$ showing that $\gamma'$ touches $AD$ at $I$ .

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