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Korean Mathematical Olympiad Final Round

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it South Korea

For a trapezoid $ABCD$ with $AB // CD$ , suppose that $A$ , $B$ , $C$ , $D$ lie in the clockwise direction. Let $\Gamma_1$ be the circle centered at $A$ , and passing through $B$ . Let $\Gamma_2$ be the circle centered at $C$ , and passing through $D$ . Let $P$ be the intersection (distinct from $B$ , $D$ ) of the line $BD$ and the circle $\Gamma_1$ . Let $\Gamma$ be the circle with the diameter $PD$ . Let $X$ be the intersection (distinct from $P$ ) of $\Gamma$ and $\Gamma_1$ . Let $Y$ be the intersection (distinct from $D$ ) of $\Gamma$ and $\Gamma_2$ . Let $Q$ be the intersection of $\Gamma_2$ and the circumcircle of $XBY$ . Show that $B$ , $D$ , $Q$ are collinear.

Solution

It suffices to show that $X$ , $B$ , $Y$ , $Q$ are cyclic under the assumption that $Q$ is the intersection of the line $BD$ and $\Gamma_2$ . There are six cases depending on the ordering of $P$ , $D$ , $B$ , $Q$ . Although we should consider all those six cases, we here give the proof only for the case where the order is $P$ , $D$ , $B$ , $Q$ .

Let $\angle ABP = \alpha$ , $\angle DPX = \beta$ . Since $AB // CD$ , we have $\angle ABP = \angle BDC = \alpha$ . For the circle $\Gamma_1$ , the circumference angle for $PB$ equals $\frac{\pi}{2} - \alpha$ ; hence we have $(\frac{\pi}{2} - \alpha) + \angle PXB = \pi$ . That is, $\angle PXB = \frac{\pi}{2} + \alpha$ .

On the other hand, we have
$\angle DYQ = \frac{1}{2} \angle DCQ = \frac{\pi}{2} - \alpha.$
For the circle $\Gamma$ , $\angle DPX = \angle DYX = \beta$ . Now it suffices to show that $\angle DBX = \angle XYQ$ .

For the triangle $DBX$ , we have
$\angle DBX = \pi - \beta - (\frac{\pi}{2} + \alpha) = \frac{\pi}{2} - \alpha - \beta.$
Meanwhile, since $\angle DYQ = \frac{\pi}{2} - \alpha$ and $\angle DYX = \beta$ , we have

$\angle$ XYQ = $\angle$ DYQ - $\angle$ DYX = $\frac{\pi}{2} - \alpha - \beta. \quad \square$

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