IMO-Auswahlklausur

Geometry Difficulty 8.0 National Olympiad, round 2 Prove it Germany

Problem:

Given is a (general) trapezoid $ABCD$ , whose diagonals intersect at the point $P$ . A point $Q$ lies between the parallels $AB$ and $DC$ in such a way that $\text{AQB = CQD}$ holds and the line $BC$ runs between $P$ and $Q$ .
Prove that $\text{DQP = BAQ}$ holds.

Solution

Solution:

Since $AB \parallel CD$ , the central dilation $\sigma$ at the intersection point $P$ of the diagonals with scale factor $k = -\frac{|CD|}{|AB|} = -\frac{|CP|}{|AP|} = -\frac{|DP|}{|BP|}$ maps the point $C$ to $A$ and $D$ to $B$ . Let $Q'$ be the image point of $Q$ under $\boldsymbol{\sigma}$ . Because

the triangle $DQC$ is mapped under $\sigma$ onto the triangle $BQ'A$ , both triangles are similar, and by the hypothesis we have $\text{AQ'B = CQD = AQB}$ . Thus $A, B, Q$ and $Q'$ lie on a circle and it holds that $\text{BAQ = BQ'Q}$ . The triangle $BQ'Q$ is mapped under $\sigma$ onto the triangle $DQQ'$ . Therefore these triangles are similar, and since $P, Q$ and $Q'$ lie on a line, it follows that $\text{BQ'Q = DQQ' = DQP}$ .

Thus $\text{BAQ = BQ'Q = DQP}$ , which was to be proved.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from de; metadata (topic, difficulty) added by this project.