JBMO Shortlist

Geometry Difficulty 5.0 AIME, harder Prove it JBMO

Problem:

Let $AB$ and $CD$ be chords in a circle of center $O$ with $A, B, C, D$ distinct, and let the lines $AB$ and $CD$ meet at a right angle at point $E$ . Let also $M$ and $N$ be the midpoints of $AC$ and $BD$ respectively. If $MN \perp OE$ , prove that $AD \parallel BC$ .

Solution

Solution:

$E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A, B, C, D$ are distinct. Let lines $AC$ and $NE$ meet at point $P$ . Then $EN = DN = BN$ (median in a right triangle), so $\angle PEC = \angle NED = \angle NDE = \angle BDC = \angle BAC = \angle EAP$ . Now $AB \perp CD$ so $EN \perp AC$ . But $OM \perp AC$ so $OM \parallel EN$ . Similarly $ON \parallel EM$ so $NEMO$ is a parallelogram (possibly degenerated). As $MN \perp OE$ , this parallelogram is a rhombus. Then the chords $AC$ and $BD$ , being equidistant from $O$ , are equal. Hence their minor arcs are equal, which means that either $AD \parallel BC$ or $AB \parallel CD$ ; the latter contradicts the fact that $AB$ and $CD$ meet at $E$ .

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