Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Ukraine

Line $l$ is perpendicular to the side $AC$ of acute triangle $ABC$ , and intersects $AC$ at $K$ . $l$ intersects circumscribed circle of $\triangle ABC$ at $P$ and $T$ (point $P$ in the same half-plane w.r.t. $AC$ as vertex $B$ ). By $P_1$ and $T_1$ are denoted projections of points $P$ and $T$ to line $AB$ . Furthermore, vertices $A, B$ belong to the segment $P_1T_1$ . Prove that the center of circumscribed circle of $\triangle P_1KT_1$ lies on the line, which contains midsegment of $\triangle ABC$ , parallel to the side $AC$ .
(Anton Tryhub)

Fig. 31

Solution

We denote by $B_1$ a projection of vertex $B$ to line $l$ , then quadrilaterals $BB_1PP_1$ and $TKAT_1$ are inscribed, with diameters $BP$ and $AT$ . Then, we have the following equalities of angles:

$\angle P_1B_1K = \pi - \angle P_1B_1P = \pi - \angle P_1BP = \\ \angle ABP = \pi - \angle ATP = \pi - \angle ATK = \\ = \pi - \angle AT_1K = \pi - \angle P_1T_1K.$

Therefore, quadrilateral $P_1B_1KT_1$ is inscribed, hence, center of circumscribed circle of $\triangle P_1KT_1$ lies on the perpendicular bisector to the segment $B_1K$ . And, clearly, midsegment of $\triangle ABC$ parallel to $AC$ lies on this exact perpendicular bisector. Q.E.D.

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