Track / Stage 6 / 193 of 400 #1193 of 1964

Problem 1193

National olympiad, first round

Geometry Difficulty 6.2 Prove it

8. (9) $A B C$ is an isosceles triangle $(A B=A C)$ . On the extensions of sides $B C, A B$ , and $A C$ , points $P, X, Y$ are chosen such that $P X \| A C$ and $P Y \| A B$ and point $P$ lies on the ray $C B$ . Point $T$ is the midpoint of the arc $B C$ of the circumcircle of triangle $A B C (T \neq A)$ . Prove that $P T \perp X Y$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution: Let $O$ be the center of the circumscribed circle of triangle $ABC$ . Then, because triangle $ABC$ is isosceles, it follows that $O$ is the midpoint of $AT$ and $\angle XBO = 180 - \angle OBA = 180 - \angle OAB = 180 - \angle OAC = \angle OAY$ . $XPYA$ is a parallelogram, from which it follows that $R = XY \cap PA$ is the midpoint of $XY$ and $PA$ , and also that $YA = PX = PB$ (the second equality follows from the fact that triangle $PXB$ is isosceles). Therefore, triangles $XBO$ and $OAY$ are congruent by two sides and the included angle ( $OA = OB$ , $BX = AY$ , $\angle OBX = \angle OAY$ ). It follows that $OR$ is the perpendicular bisector in the isosceles triangle $YOX$ . Moreover, $OR$ is the midline of triangle $APT$ . Thus, $PT \perp XY$ follows from the fact that $OR \parallel PT$ and $OR \perp XY$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.