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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

Point $P$ is chosen on the smaller arc $BC$ of the circumscribed circle of an acute-angled triangle $ABC$ . Points $R$ and $S$ on sides $AB$ and $AC$ respectively are chosen such that $CPRS$ is a parallelogram. Point $T$ on arc $AC$ of the circumscribed circle of $\triangle ABC$ is such that $BT \parallel CP$ . Prove that $\angle TSC = \angle BAC$ .

(Anton Tryhub)

Solution

Let line $PR$ intersect the circumscribed circle of $\triangle ABC$ at point $K$ (Fig. 27).

Then, $AKPC$ and $BPCT$ are isosceles trapezoids, hence, $\angle RSA = \angle PCA = 180^\circ - \angle RKA$ , so $AKRS$ is inscribed, which makes it an isosceles trapezoid.

Then, $\angle RKS = \angle RAS = \angle BAC = \angle PKT = \angle RKT$ . Hence, points $K$ , $S$ , $T$ lie on the same line.

Then, $\angle TSC = \angle TKP = \angle BAC$ .

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