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Auswahlwettbewerb zur Internationalen Mathematik-Olympiade

Geometry Difficulty 8.3 Shortlist Prove it Germany

Problem:

Let $ABCD$ be a convex parallelogram that is acute-angled at $A$ . The reflections of $A$ across the lines $BC$ and $CD$ are denoted by $P$ and $Q$ , respectively. Moreover, the line $BD$ intersects the segments $\overline{AP}$ and $\overline{AQ}$ in their interior at the points $R$ and $S$ , respectively.

Prove that the circumcircles of the triangles $BRP$ and $DQS$ are tangent to each other.

Solution

Solution:

We consider the reflection $Z$ of $A$ under reflection across $BD$ . We will prove that the two circles mentioned in the problem statement are tangent to each other at $Z$ .

First, because of $\measuredangle BZR = \measuredangle RAB = \measuredangle BPR$ and the converse of the inscribed angle theorem, it is clear that $Z$ lies on the circumcircle $\omega_1$ of $BRP$ . Analogously, one shows that $Z$ also lies on the circumcircle $\omega_2$ of $DQS$ . The tangent-chord angle of the chord $ZB$ of $\omega_1$ is $\angle ZRB = \measuredangle BRA$ . Since $AR$ is perpendicular to $BC$ and hence also to $AD$ , this angle has the size $90^\circ - \angle ADB$ . Analogously, one shows that the tangent-chord angle of the chord $ZD$ of $\omega_2$ equals $90^\circ - \angle DBA$ . The sum of these two angles is now $180^\circ - \measuredangle ADB - \angle DBA = \angle BAD = \angle DZB$ , which shows that the tangents to $\omega_1$ and $\omega_2$ at the point $Z$ coincide. Thus the two circles are indeed tangent to each other.

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