IMO 1973 SL 2

Origin: CZS

Problem

Given a circle K, find the locus of vertices A of parallelograms ABCD with diagonals AC \leqBD, such that BD is inside K.

Solution

Let D′ be the reflection of D across A. Since BCAD′ is then a parallel- ogram, the condition BD \geqAC is equivalent to BD \geqBD′, which is in turn equivalent to \angleBAD \geq\angleBAD′, i.e. to \angleBAD \geq90◦. Thus the needed locus is actually the locus of points A for which there exist points B, D inside K with \angleBAD = 90◦. Such points B, D exist if and only if the two tangents from A to K, say AP and AQ, determine an obtuse angle. Then if P, Q \inK, we have \anglePAO = \angleQAO = ϕ > 45◦; hence OA = OP sin ϕ < OP \sqrt 2. Therefore the locus of A is the interior of the circle K′ with center O and radius \sqrt 2 times the radius of K.