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Problem 1819

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it IMO HK TST · Hong Kong

Let $D$ be an arbitrary point inside $\triangle ABC$ . Let $\Gamma$ be the circumcircle of $\triangle BCD$ . The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$ . The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$ . The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\triangle BFP$ and $\triangle CEQ$ always pass through the same fixed point regardless of the position of $D$ . (Assume all the labelled points are distinct.)

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

Let $BE$ and $CF$ intersect at the $A$ -excentre $J$ of $\triangle ABC$ . We claim that $J$ is the desired fixed point.

We only consider the configuration as shown since the other cases are similar. Since $BE$ is the external angle bisector, we have $\angle PBE = \angle EBC$ . This is equal to $\angle EFJ$ as $B, E, F, C$ are concyclic. This implies $P, B, F, J$ are concyclic. Similarly, $Q, C, E, J$ are concyclic. This shows the two circles always pass through $J$ , which is independent of $D$ .

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