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Problem 1811

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it Baltic Way 2021 Shortlist · Baltic Way · 2021

Let points $A$ and $B$ lie on circle $\omega$ with center $O$ . Assume that $O$ does not lie on line $AB$ . Let point $C$ lie on segment $AB$ and denote by $M$ and $N$ the midpoints of segments $AC$ and $CB$ , respectively. The circumcircle of $AON$ intersects $\omega$ at $A$ and $K$ and the circumcircle of $BOM$ intersects $\omega$ at $B$ and $L$ . Moreover, circumcircles of $AON$ and $BOM$ intersect each other at $O$ and $X$ . Prove that the quadrilateral $CKXL$ is cyclic.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution. Let $K'$ and $H'$ be the projections of $P'$ onto $AB$ and $AC$ respectively, as in figure 14. Now, $HPP'H'$ is a right angled trapezium, and $M$ is the midpoint of $PP'$ . If $M'$ is the midpoint of $HH'$ , then $M'M||HP$ , so $M'M \perp HH'$ . Therefore, $MH = MH'$ . Similarly, $MK = MK'$ , and so as $MK = MH, K', K, H, H'$ lie on a circle with centre $M$ .
As $KHH'K'$ is cyclic, we have $\angle H'K'A = \angle KHA$ . By considering right angles, we see that $AH'P'K'$ and $AHPK$ are cyclic.
We get
$\begin{align*} \angle PAB &= \angle PAK \\ &= \angle PHK \\ &= 90^\circ - \angle KHA \\ &= 90^\circ - \angle H'K'A \\ &= \angle P'K'H' \\ &= \angle P'AH' \\ &= \angle P'AC \end{align*}$
as required.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.