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, 2015

Geometry Difficulty 5.3 AIME, harder Prove it Taiwan

Let $ABC$ be a triangle with incircle $\omega$ , incentre $I$ and circumcircle $\Gamma$ . Let $D$ be the tangency point of $\omega$ with $BC$ , let $M$ be the midpoint of $ID$ , and let $A'$ be the diametral opposite of $A$ with respect to $\Gamma$ . If we denote $X = A'M \cap \Gamma$ then prove that the circumcircle of $\triangle AXD$ is tangent to $BC$ .

Solution

Consider a circle passing through $A$ and tangent to line $BC$ at point $D$ ; let its second intersection with $\Gamma$ be $Y$ . We shall prove that $Y, M, A'$ are collinear. Then $X = Y$ , and the problem is solved.

Let $\omega$ touch $AC, AB$ at points $E, F$ respectively; let $EF$ meet $BC$ at point $P$ , and let $AX$ meet $BC$ at point $J$ .

Since $JB \cdot JC = JA \cdot JY = JD^2$ and $(B, C, D, P) = -1$ , $J$ is the midpoint of $DP$ . Note that $AD$ is the polar of point $P$ with respect to circle $\omega$ , so $AD$ is perpendicular to $PI$ ; let the foot of perpendicular be point $U$ . Therefore, the midline $JM$ of the two sides of $\triangle PID$ is the perpendicular bisector of $UD$ . Hence $\angle JUD = \angle JDU = \angle AXD$ , so $J, Y, U, M, D$ are concyclic. From this we get $\angle AYM = \angle JDM = 90^\circ$ , so $YM$ meets $\Gamma$ at the diametral opposite $A'$ of $A$ , i.e., $Y, M, A'$ are collinear. This completes the proof.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.