Competition problem — Geometry — Maths Olympiad Prep

Library / /10 of 35

Geometry Difficulty 5.1 AIME, harder Prove it Belarus

An arbitrary point $D$ is marked inside the triangle $ABC$ . The lines $AD$ , $BD$ and $CD$ intersect the sides $BC$ , $CA$ and $AB$ at the points $K$ , $L$ and $M$ respectively. Let $A_1$ , $B_1$ and $C_1$ be the midpoints of the segments $BC$ , $CA$ and $AB$ , and $X$ , $Y$ be the midpoints of the segments $ML$ and $KM$ respectively. The lines $XA_1$ and $YB_1$ intersect at the point $F$ .
Prove that $FO \parallel DH$ , where $O$ is the circumcenter of the triangle $A_1B_1C_1$ and $H$ is the orthocenter of the triangle $ABC$ .
(Mikhail Karpuk)

Solution

Note that $XA_1$ is the Gauss-Newton line of a complete quadrangle formed by the lines $BL$ , $BA$ and $CM$ , $CA$ , hence it passes through the midpoint of the segment $AD$ . Similarly $YB_1$ passes through the midpoint of the segment $BD$ . Therefore $F$ is the center of mass of points $A$ , $B$ , $C$ and $D$ with unit masses. This implies that $F$ lies on the line $DG$ and $GF: FD = 1:3$ where $G$ is the centroid of the triangle $ABC$ . At the same time $O$ is the nine-point center of the triangle $ABC$ so it also lie on the Euler line of the triangle $ABC$ and $GO: OH = 1:3$ . Therefore the Thales's theorem implies that $FO \parallel DH$ .

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.