Autumn Tournament

Geometry Difficulty 5.6 AIME, harder Prove it Bulgaria

Given an acute triangle $ABC$ with circumcenter $O$ . The point $P$ on $BC$ such that $BP < \frac{BC}{2}$ and the point $Q$ is on $BC$ , such that $CQ = BP$ . The line $AO$ meets $BC$ at $D$ and $N$ is the midpoint of $AP$ . The circumcircle of $(ODQ)$ meets $(BOC)$ at $E$ . The lines $NO$ , $OE$ meet $BC$ at $K$ , $F$ . Show that $AOKF$ is cyclic.

(Alexander Ivanov)

Solution

Let $A'$ be the antipode of $A$ and let $AO \cap (BOC) = R$ . Since $NO \parallel PA'$ , by Reim's theorem we need $APA'F$ being cyclic, or $DP \cdot DF = DA \cdot DA' = DB \cdot DC = DO \cdot DR$ , so we need $OPRF$ being cyclic, or that $\angle OFP = \angle ORP$ . By shooting lemma, $RDEF$ is cyclic, so $\angle OFP = \angle ERD$ , so we need $AR$ being the angle bisector of $\angle PRE$ . Since $AR$ bisects $\angle BRC$ , it is sufficient to show that $\angle BRP = \angle CRE$ .

To use that $EQDO$ is cyclic, let $EQ \cap (BOC) = S$ ; Reim's implies $RS \parallel BC$ , which together with the length condition gives that $RSQP$ is an isosceles trapezoid. Hence, $\angle CRE = \angle CSQ = \angle BRP$ , which finishes the problem.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.