Competition problem — Geometry — Maths Olympiad Prep

Library / /7 of 13

Geometry Difficulty 6.3 National olympiad Prove it Bulgaria

In a triangle $\triangle ABC$ points $L$ , $P$ and $Q$ lie on the segments $AB$ , $AC$ and $BC$ , respectively, and are such that $PCQL$ is a parallelogram. The circle with center the midpoint $M$ of the segment $AB$ and radius $CM$ and the circle of diameter $CL$ intersect for the second time at point $T$ . Prove that the lines $AQ$ , $BP$ and $LT$ intersect in a point.

Solution

Since $AC \parallel LQ$ and $BC \parallel LP$ , we have $S_{ALQ} = S_{CLQ} = S_{PLC} = S_{PLB}$ . Let the point $K$ be such that $AKBC$ is a parallelogram. By analogy we have $S_{AKQ} = S_{AKC} = S_{CKB} = S_{PKB}$ .

The locus of the points $X$ such that $\triangle AXQ$ and $\triangle PXB$ are oriented in one and the same direction and have equal areas is a straight line $\ell$ passing through the intersecting point of $AQ$ and $BP$ . Therefore $\ell \equiv KL$ and $AQ$ , $BP$ and $KL$ intersect in a point.

Let $N$ be the midpoint of $CL$ . The line $KL$ is the image of $MN$ under homothety of center $C$ and coefficient $2$ . Since the point $T$ is symmetric to $C$ with respect to $MN$ we have that $T$ lies on $KL$ which completes the proof.

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 and solution reproduced as published; topic and difficulty added by this site.