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Problem 1294

AIME late

Geometry Difficulty 5.0 Prove it Taiwan IMO Selection Camp · Taiwan · 2023

In the plane, let point $O$ be the center of circle $\Gamma$ , and let there be two points $A$ , $B$ on $\Gamma$ such that $O, A, B$ are not collinear. Let point $M$ be the midpoint of segment $AB$ , and take points $P$ , $Q$ on lines $OA$ , $OB$ respectively such that $P \neq A$ and points $P, M, Q$ are collinear. Let the line through $P$ parallel to $AB$ and the line through $Q$ parallel to $OM$ meet at point $X$ ; further let the line through $X$ parallel to $OA$ and the line through $B$ perpendicular to $OX$ meet at point $Y$ . Prove that: if point $X$ lies on circle $\Gamma$ , then point $Y$ also lies on circle $\Gamma$ .

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

Without loss of generality, let $OA = OB = OX = 1$ . Let the directed angles $\angle AOB = \theta$ , $\angle BOX = \varphi$ . Also let $A = \varphi + \theta/2$ , $B = \theta/2$ . Applying the Law of Sines in $\triangle OPX$ , we find that the directed length $OP$ equals $\cos A/ \cos B$ . Also applying the Law of Sines in $\triangle OQX$ , we find that the directed length $OQ$ equals $\sin A/ \sin B$ . Then by Menelaus's theorem,
$\frac{\frac{\cos A}{\cos B}}{1 - \frac{\cos A}{\cos B}} = - \frac{\frac{\sin A}{\sin B}}{1 - \frac{\sin A}{\sin B}}$
Expanding the above expression gives
$\cos A \sin B + \sin A \cos B - 2 \cos A \sin A = 0,$
that is, $\sin(A+B) = \sin 2A$ . From this we deduce that $A+B \equiv 2A \pmod{2\pi}$ or $3A+B \equiv \pi \pmod{2\pi}$ .
In the first case, we obtain $A=B$ , so $\varphi=0$ . This means $X=B$ and $Y=B$ lie on $\Gamma$ .
In the second case, we have $3\varphi \equiv \pi - 2\theta \pmod{2\pi}$ . Note that if the line through $B$ perpendicular to $OX$ meets $\Gamma$ at point $Y'$ , then $\angle XOY' = \varphi$ . Then $\angle AOX + \angle AOY' = 3\varphi + 2\theta = \pi$ , so we know that $AO$ is parallel to $XY'$ , from which it follows that $Y=Y'$ lies on $\Gamma$ . This completes the proof.

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