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Problem 2318

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.8 Prove it International Mathematical Olympiad · China

In triangle $ABC$ the bisector of angle $BCA$ intersects the circumcircle at $R$ , the perpendicular bisector of $BC$ at $P$ , and the perpendicular bisector of $AC$ at $Q$ . The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$ . Prove that the triangles $RPK$ and $RQL$ have the same area.

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

If $AC = BC$ , $\triangle ABC$ is an isosceles triangle, and $CR$ is the symmetry axis of $\triangle RQL$ and $\triangle RPK$ . The conclusion is obviously true.

If $AC \ne BC$ , without loss of generality, let $AC < BC$ . Denote the center of circumcircle of $\triangle ABC$ by $O$ .

Since the right triangles $CQL$ and $CPK$ are similar,

$\angle CPK = \angle CQL = \angle OQP, \text{ and } \frac{QL}{PK} = \frac{CQ}{CP}. \quad ①$
Let $l$ be the perpendicular bisector of $CR$ , then $O$ is on $l$ .
Since $\triangle OPQ$ is an isosceles triangle, $P$ and $Q$ are two points symmetrical about $l$ on $CR$ .
So
$RP = CQ \quad \text{and} \quad RQ = CP. \qquad \textcircled{2}$
By ①, ②,
$\begin{aligned} \frac{S(\triangle RQL)}{S(\triangle RPK)} &= \frac{\frac{1}{2} \cdot RQ \cdot QL \cdot \sin \angle RQL}{\frac{1}{2} \cdot RP \cdot PK \cdot \sin \angle RPK} \\ &= \frac{RQ}{RP} \cdot \frac{QL}{PK} = \frac{CP}{CQ} \cdot \frac{CQ}{CP} = 1. \end{aligned}$
Hence the two triangles have the same area.

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