Track / Stage 6 / 89 of 400 #1729 of 2604

Problem 1729

National Olympiad, first round

Geometry Difficulty 6.1 Prove it Singapore Mathematical Olympiad (SMO) · Singapore

Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$ draw a line meeting the side $AB$ at $P$ and the side $CD$ at $Q$ . Find all the quadrilaterals so that there exists the segment $PQ$ that divides the triangles $ABM$ and $CDM$ into 4 similar triangles.

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

Suppose $\angle APM > 90^\circ$ . Then in $\triangle BPM$ , $\angle BPM < 90^\circ$ . Since $\angle PBM + \angle BMP = \angle APM$ , $\angle PBM, \angle BMP < \angle APM$ . So none of the angles of $\triangle BPM$ can be equal to $\angle APM$ of $\triangle APM$ . Therefore $\triangle APM$ cannot be similar to $\triangle BPM$ . Thus $\angle APM = 90^\circ$ and $PQ \perp AB$ . Similarly $PQ \perp CD$ and it follows that $AB \parallel DC$ .

It then follows that $\triangle APM \sim \triangle CQM$ and $\triangle BMP \sim \triangle DQM$ .
If $\triangle APM \sim \triangle BPM$ , then $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$ . Therefore $ABCD$ is an isosceles trapezium.
If $\triangle APM \sim \triangle MPB$ , then $\angle MAP = \angle BMP$ and $\angle PMA = \angle PBM$ . It follows that $\angle AMB = 90^\circ$ . So the diagonals are perpendicular. The quadrilaterals are either isosceles trapezia or trapezia in which the diagonals are perpendicular.

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