Competition problem — Geometry — Maths Olympiad Prep

Library / /14 of 87

Geometry Difficulty 5.5 AIME, harder Prove it Serbia

Problem:

On the sides $AB$ , $AC$ and $BC$ of triangle $ABC$ points $M$ , $X$ and $Y$ are given, respectively, such that $AX = MX$ and $BY = MY$ . Let $K$ and $L$ be, respectively, the midpoints of segments $AY$ and $BX$ , and let $O$ be the center of the circumscribed circle of triangle $ABC$ . If $O_{1}$ and $O_{2}$ are the points symmetric to point $O$ with respect to $K$ and $L$ , respectively, prove that the points $X$ , $Y$ , $O_{1}$ and $O_{2}$ lie on the same circle.

Solution

Solution:

Let us set up a coordinate system with the origin at point $M$ and the $x$ -axis along the line $AB$ . Let points $X$ and $Y$ have coordinates $(a, b)$ and $(c, d)$ respectively. Because $AX = XM$ and $BY = YM$ , the coordinates of points $A$ and $B$ are $(2a, 0)$ and $(2c, 0)$ , and those of points $K$ and $L$ are $\left(a + \frac{c}{2}, \frac{d}{2}\right)$ and $\left(c + \frac{a}{2}, \frac{b}{2}\right)$ , respectively. Point $O$ has coordinates $(a + c, e)$ for some $e$ , from which we obtain $O_{1}(a, d - e)$ and $O_{2}(c, b - e)$ . Therefore, points $O_{1}$ and $O_{2}$ are symmetric to points $X$ and $Y$ with respect to the line $y = \frac{b + d - e}{2}$ , so $X$ , $Y$ and $O_{1}$ , $O_{2}$ are vertices of a (possibly degenerate) isosceles trapezoid, and therefore lie on a circle.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from sr; metadata (topic, difficulty) added by this project.