Competition problem — Geometry — Maths Olympiad Prep

Library / /91 of 136

Geometry Difficulty 8.2 Shortlist Prove it Hong Kong

Let $ABC$ be an equilateral triangle with circumcentre $O$ . Let $P$ be a point inside the triangle such that $\angle BPC = 120^\circ$ and $P \ne O$ . Let $B'$ and $C'$ be the images of $B$ and $C$ respectively under the reflection in the line $PO$ . Furthermore $B'C'$ meets $AP$ at $D$ . Determine $AD : DP$ . Justify your claim.

Solution

We have $AD : DP = 1 : 1$ .
Since $\angle BPC = 120^\circ = \angle BOC$ , the points $B$ , $C$ , $O$ , $P$ are concyclic. As
$\angle BPO = 180^\circ - \angle OCB = 150^\circ,$
we have $\angle B'PB = 2(180^\circ - 150^\circ) = 60^\circ$ , and hence $\triangle PB'B$ is equilateral. Similarly, as
$\angle CPO = \angle CBO = 30^\circ,$
we have $\angle CPC' = 60^\circ$ , and hence $\triangle PCC'$ is equilateral.

Now, since $\triangle BPB'$ and $\triangle BCA$ are directly similar, by spiral similarity, we have $\triangle BAB' \sim \triangle BCP$ . In the same way, we have $\triangle ACC' \sim \triangle BCP$ . Therefore, we obtain $\triangle BAB' \sim \triangle ACC'$ . Indeed, they are congruent since $BA = AC$ . This implies $AC' = BB' = B'P$ and $B'A = C'C = PC'$ . Thus, $AB'PC'$ is a parallelogram. The diagonals $AP$ and $B'C'$ bisect each other. In particular, we have $AD : DP = 1 : 1$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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