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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

In the triangle $ABC$ $\angle A=75^\circ$ and $\angle C=45^\circ$ . Points $P$ and $T$ are chosen on the segments $AB$ and $BC$ in such a way that quadrilateral $APTC$ is cyclic and $CT = 2AP$ . Point $O$ is the circumcenter of $\triangle ABC$ . The ray $TO$ crosses side $AC$ in a point $K$ . Prove that $TO=OK$ .

(Anton Trygub)

Solution

Let $CD$ be a diameter of $(ABC)$ . Then $\triangle ADC$ is a right triangle with an angle $60^\circ$ . Hence, $CD = 2AD$ (fig. 30) and $\triangle DTC \sim \triangle DPA$ by two proportional sides and equal included angles. Therefore, $\angle BPD = \angle BTD$ and $PDBT$ is cyclic quadrilateral. We point out that $\angle BTD = \angle BPD = \angle DPT - \angle BPT = 180^\circ - \angle DBT - 45^\circ = 180^\circ - 90^\circ - 45^\circ = 45^\circ$ , from which $DT \parallel AC$ . Since $DO = OC$ and $DT \parallel CK$ , $DTCK$ is parallelogram and $TO = OK$ .

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