IMO 2J, Mock Exam 1

Geometry Difficulty 8.8 Shortlist Prove it Taiwan

Given triangle $ABC$ . Let $BPCQ$ be a parallelogram ( $P$ is not on $BC$ ). Let $U$ be the intersection of $CA$ and $BP$ , $V$ be the intersection of $AB$ and $CP$ , $X$ be the intersection of $CA$ and the circumcircle of triangle $ABQ$ distinct from $A$ , and $Y$ be the intersection of $AB$ and the circumcircle of triangle $ACQ$ distinct from $A$ . Prove that $\overline{BU} = \overline{CV}$ if and only if the lines $AQ$ , $BX$ , and $CY$ are concurrent.

Solution

By the radical center theorem, $AQ, BX, CY$ are concurrent if and only if $B, C, X, Y$ are concyclic, and via $\angle BXC = \angle BQA, \angle BYC = \angle AQC$ , this is in turn equivalent to $QA$ being one of the angle bisectors of $\angle BQC$ , that is, $\frac{\sin \angle BQA}{\sin \angle AQC} = \pm 1$ (note that $Q$ is not on $BC$ ).

On the other hand, by the trigonometric form of Ceva's theorem,
$\frac{\sin \angle BQA}{\sin \angle AQC} \cdot \frac{\sin \angle CBA}{\sin \angle ABQ} \cdot \frac{\sin \angle QCA}{\sin \angle ACB} = 1.$
Therefore $\frac{\sin \angle BQA}{\sin \angle AQC} = \pm 1$ if and only if $\frac{\sin \angle CBA}{\sin \angle ABQ} \cdot \frac{\sin \angle QCA}{\sin \angle ACB} = \pm 1$ . And
$\left| \frac{\sin \angle CBA}{\sin \angle ABQ} \cdot \frac{\sin \angle QCA}{\sin \angle ACB} \right| = \frac{\sin \angle CBV}{\sin \angle BVC} \cdot \frac{\sin \angle BUC}{\sin \angle UCB} = \frac{\overline{CV}}{\overline{BC}} \cdot \frac{\overline{BC}}{\overline{BU}} = \frac{\overline{CV}}{\overline{BU}}$
Hence the two conditions are equivalent.

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