Competition problem — Geometry — Maths Olympiad Prep

Library / /32 of 62

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

In the triangle $ABC$ $AB > AC$ . A tangent to the circumcircle of the triangle $ABC$ has been drawn through the point $A$ . This tangent intersects the line $BC$ at the point $P$ . In continuation of the side $BA$ after the point $A$ point $Q$ was selected such that $AQ = AC$ . Let $X$ and $Y$ be the midpoints of the segments $CQ$ and $AP$ , respectively, and let $R$ belong to the segment $AP$ such that $AR = CP$ . Prove that $CR = 2XY$ .

Solution

Consider the point $M$ , which is the midpoint of the segment $PQ$ , then $MX$ and $MY$ are the mid-segments of $\triangle CPQ$ and $\triangle APQ$ respectively. It follows that $MX = y$ and $MY = x$ , and also $MX \parallel CP$ and $MY \parallel AQ$ . This parallel gives that $\angle XMY = \angle PBQ = \beta$ (as the angles of respectively parallel sides).

Fig. 40
Thus, $\frac{MY}{MX} = \frac{x}{y} = \frac{2x}{2y} = \frac{AC}{AR}$ and $\angle XMY = \angle PAC = \beta$ . Which means $\triangle XMY \sim \triangle RAC$ , due to the proportionality of the two sides and the angle between them. From the similarity of triangles it follows that
$\frac{CR}{XY} = \frac{AC}{MY} = \frac{2x}{x} = 2.$

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 reproduced verbatim; metadata (topic, difficulty) added by this project.