Track / Stage 6 / 21 of 400 #1021 of 1964

Problem 1021

National olympiad, first round

Geometry Difficulty 6.0 Prove it

17.10. Given an acute angle $M O N$ and points $A$ and $B$ inside it. Find a point $X$ on the side $O M$ such that the triangle $X Y Z$ , where $Y$ and $Z$ are the points of intersection of the lines $X A$ and $X B$ with $O N$ , is isosceles: $X Y = X Z$ .

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

17.10. Let the projection of point $A$ onto line $O N$ be closer to point $O$ than the projection of point $B$ . Suppose that the isosceles triangle $X Y Z$ is constructed.

Fig. 17.3

Consider point $A^{\prime}$ , which is symmetric to point $A$ with respect to line $O M$ . Drop a perpendicular $X H$ from point $X$ to line $O N$ (Fig. 17.3). Since $\angle A^{\prime} X B = \angle A^{\prime} X O + \angle O X A + \angle Y X H + \angle H X Z = 2 \angle O X Y + 2 \angle Y X H = 2 \angle O X H = 180^{\circ} - 2 \angle M O N$ , the angle $\angle A^{\prime} X B$ is known. Point $X$ is the intersection of line $O M$ and the arc from which segment $A^{\prime} B$ is seen at an angle of $180^{\circ} - 2 \angle M O N$ . In this case, the projection of point $X$ onto line $O N$ must lie between the projections of points $A$ and $B$ .

Conversely, if $\angle A^{\prime} X B = 180^{\circ} - 2 \angle M O N$ and the projection of point $X$ onto line $O N$ lies between the projections of points $A$ and $B$ , then triangle $X Y Z$ is isosceles.

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