Track / Stage 7 / 7 of 300 #1407 of 1964

Problem 1407

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.0 Prove it Indija TS 2012 · India · 2012

The circumcenter of the cyclic quadrilateral $ABCD$ is $O$ . The second intersection point of the circles $ABO$ and $CDO$ , other than $O$ , is $P$ , which lies in the interior of the triangle $DAO$ . Choose a point $Q$ on the extension of $OP$ beyond $P$ , and a point $R$ on the extension of $OP$ beyond $O$ . Prove that $\angle QAP = \angle OBR$ holds if and only if $\angle PDQ = \angle RCO$ .

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 $H$ be the radical center of the circles $ABCD$ , $ABOP$ and $CDPO$ . Then the radical axes of any two of these circles, i.e. the lines $AB$ , $CD$ and $OP$ , pass through $H$ . Since $P$ lies on the shorter arcs $AO$ and $DO$ , it follows that $H$ lies on the extension of $OP$ beyond $P$ . The radical center satisfies
$HA \cdot HB = HC \cdot HD = HO \cdot HP.$
(1)
Since the quadrilateral $ABOP$ is cyclic,

$\begin{aligned} \angle QAB + \angle BRQ &= (\angle PAB + \angle QAP) + (\angle BOP - \angle OBR) \\ &= (\angle PAB + \angle BOP) + (\angle QAP - \angle OBR) \\ &= 180^\circ + (\angle QAP - \angle OBR). \end{aligned}$
Therefore, $\angle QAP = \angle OBR$ holds if and only if the quadrilateral $ABRQ$ is cyclic, which is equivalent to $HQ \cdot HR = HA \cdot HB$ .
Similarly, $\angle QAP = \angle OBR$ holds if and only if $HQ \cdot HR = HC \cdot HD$ .
Combining with (1),
$\angle QAP = \angle OBR \Leftrightarrow HQ \cdot HR = HA \cdot HB \Leftrightarrow HQ \cdot HR = HC \cdot HD \Leftrightarrow \angle PDQ = \angle RCO.$

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