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Problem 881

AMC 12 late, AIME early

Geometry Difficulty 4.6 Prove it Berkeley Math Circle · United States

Let $M$ be an interior point of a parallelogram $ABCD$ . Prove that $MA + MB + MC + MD$ is strictly less than the length of the perimeter of $ABCD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Denote by $X$ and $Y$ the points of intersection of the segments $AB$ and $CD$ with the line through $M$ parallel to $BC$ . Similarly, let $U$ and $V$ denote the points of intersection of the segments $AD$ and $BC$ with the line through $M$ parallel to $AB$ .

Then $MA < AU + UM = XM + UM$ , $MB < MX + XB = MX + MV$ , $MC < MV + VC = MV + MY$ , and $MD < MY + YD = MY + MU$ .

Hence
$MA + MB + MC + MD < 2(XM + YM) + 2(UM + VM) = 2AD + 2AB = AB + BC + CD + DA.$

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