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Problem 2449

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it Team Selection Test for IMO · Turkey · 2010

For an interior point $D$ of a triangle $ABC$ , let $\Gamma_D$ denote the circle passing through the points $A$ , $E$ , $D$ , $F$ if these points are concyclic where $BD \cap AC = \{E\}$ and $CD \cap AB = \{F\}$ . Show that all circles $\Gamma_D$ pass through a second common point different from $A$ as $D$ varies.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $A'$ be the midpoint of the side $BC$ , and let $A''$ be the second point of intersection of the line $AA'$ and the circumcircle $\Gamma$ of the triangle $BDC$ . Since both $D, E, A, F$ and $D, C, A'', B$ are concyclic, $\angle BA''C = \angle BAC$ . Hence $A'$ is the midpoint of $AA''$ . In particular, the circle $\Gamma$ and $J$ , the second point of intersection of $\Gamma$ and $AA''$ , do not depend on the point $D$ . We will show that $J$ lies on $\Gamma_D$ .
If $D$ lies on the same side of $AA'$ as $B$ , then we have $\angle JCB = \angle JDE$ by concyclicity of $B, D, J, C$ . Considering the power of $A'$ with respect to $\Gamma$ we obtain $A'J \cdot A'A'' = A'B \cdot A'C$ , hence $A'J \cdot A'A = A'C^2$ implying that the line $A'C$ is tangent to the circumcircle of the triangle $AJC$ . In particular, we have $\angle JCB = \angle JAC$ . Therefore $\angle JDE = \angle JAE$ , and $J$ lies on $\Gamma_D$ . A similar reasoning works if $D$ lies on the same side of $AA'$ as $C$ .

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