Library / Canadian Mathematical Olympiad /1 of 57

Canadian Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it Canada

Problem:

$ABCD$ is a convex quadrilateral in which $AB$ is the longest side. Points $M$ and $N$ are located on sides $AB$ and $BC$ respectively, so that each of the segments $AN$ and $CM$ divides the quadrilateral into two parts of equal area. Prove that the segment $MN$ bisects the diagonal $BD$ .

Solution

Solution:

Since $[MADC] = \frac{1}{2}[ABCD] = [NADC]$ , it follows that $[ANC] = [AMC]$ , so that $MN \parallel AC$ . Let $m$ be a line through $D$ parallel to $AC$ and $MN$ and let $BA$ produced meet $m$ at $P$ and $BC$ produced meet $m$ at $Q$ . Then
$[MPC] = [MAC] + [CAP] = [MAC] + [CAD] = [MADC] = [BMC]$
whence $BM = MP$ . Similarly $BN = NQ$ , so that $MN$ is a midline of triangle $BPQ$ and must bisect $BD$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.