Track / Stage 6 / 105 of 400 #1105 of 2000

Problem 1105

National olympiad, first round

Geometry Difficulty 6.1 Prove it XXIII OBM · Brazil

In a convex quadrilateral, the *altitude* relative to a side is defined to be the line perpendicular to this side through the midpoint of the opposite side. Prove that the four altitudes have a common point if and only if the quadrilateral is cyclic, that is, if and only if, there exists a circle which contains its four vertices.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Consider the following
Lemma. Let $ABCD$ be a convex quadrilateral. The lines that connect the midpoints of opposite sides meet in $E$ and the perpendicular bisectors of opposite sides $AB$ and $CD$ meet in $O$ . Then the altitudes relative to $AB$ and $CD$ meet in a point $O'$ symmetric to $O$ with respect to $E$ .

*Proof.* It's well known that the quadrilateral whose vertices are the midpoints of $ABCD$ is a parallelogram. So $E$ is the midpoint of both its diagonals, in particular $MN$ . But since the pairs of lines $OM$ , $NO'$ and $ON$ , $MO'$ are parallel, $MO'NO$ is also a parallelogram, so the diagonals $MN$ and $OO'$ meet in their respective midpoints. This means that $E$ is the midpoint of $OO'$ and we're done.

Let $O_1$ and $O_2$ be the intersection of the perpendicular bisectors of $AB$ , $CD$ and $AD$ , $BC$ , respectively and $O_1'$ and $O_2'$ be the intersection points of the altitudes relative to $AB$ , $CD$ and $AD$ , $BC$ respectively. The four altitudes have a common point if and only if $O_1' = O_2'$ , which by the lemma is equivalent to $O_1 = O_2$ , which is, in turn, equivalent to $ABCD$ being cyclic.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.