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Problem 2505

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.7 Prove it IMO Shortlist · IMO

Let $AB C$ be a triangle with $AB \neq AC$ and circumcenter $O$ . The bisector of $\angle BAC$ intersects $BC$ at $D$ . Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$ . The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BX CY$ is cyclic.

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

The bisector of $\angle BAC$ and the perpendicular bisector of $BC$ meet at $P$ , the midpoint of the minor arc $\widehat{BC}$ (they are different lines as $AB \neq AC$ ). In particular $OP$ is perpendicular to $BC$ and intersects it at $M$ , the midpoint of $BC$ .

Denote by $Y'$ the reflection of $Y$ with respect to $OP$ . Since $\angle BYC = \angle BY'C$ , it suffices to prove that $BX CY'$ is cyclic.

We have
$\angle XAP = \angle OPA = \angle EYP.$
The first equality holds because $OA = OP$ , and the second one because $EY$ and $OP$ are both perpendicular to $BC$ and hence parallel. But $\{Y, Y'\}$ and $\{E, D\}$ are pairs of symmetric points with respect to $OP$ , it follows that $\angle EYP = \angle DY'P$ and hence
$\angle XAP = \angle DY'P = \angle XY'P.$
The last equation implies that $XAY'P$ is cyclic. By the powers of $D$ with respect to the circles $(XAY'P)$ and $(ABPC)$ we obtain
$XD \cdot DY' = AD \cdot DP = BD \cdot DC$
It follows that $BX CY'$ is cyclic, as desired.

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