IMO 1989 SL 14

Origin: IND

Problem

A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.

Solution

Lemma 1. In a quadrilateral ABCD circumscribed about a circle, with points of tangency P, Q, R, S on DA, AB, BC, CD respectively, the lines AC, BD, PR, QS concur. Proof. Follows immediately, for example, from Brianchon’s theorem. Lemma 2. Let a variable chord XY of a circle C(I, r) subtend a right angle at a fixed point Z within the circle. Then the locus of the midpoint P of XY is a circle whose center is at the midpoint M of IZ and whose radius is

r2/2 −IZ2/4. Proof. From \angleXZY = 90◦follows −−\to ZX \cdot−−\to ZY = (−\to IX −−\to IZ)\cdot(−\to IY −−\to IZ) = 0. Therefore, −−\to MP 2 = (−−\to MI + −\to IP)2 = 1 4(−−\to IZ + −\to IX + −\to IY )2 = 1 4(IX2 + IY 2 −IZ2 + 2(−\to IX −−\to IZ) \cdot (−\to IY −−\to IZ)) = 1 2r2 −1 4IZ2. Lemma 3. Using notation as in Lemma 1, if ABCD is cyclic, PR is perpendicular to QS. Proof. Consider the inversion in C(I, r), mapping A to A′ etc. (P, Q, R, S are fixed). As is easily seen, A′, B′, C′, D′ will lie at the midpoints of PQ, QR, RS, SP, respectively. A′B′C′D′ is a parallelogram, but also cyclic, since inversion preserves circles; thus it must be a rectangle, and so PR \perpQS. Now we return to the main result. Let I and O be the incenter and circum- center, Z the intersection of the diagonals, and P, Q, R, S, A′, B′, C′, D′ points as defined in Lemmas 1 and 3. From Lemma 3, the chords PQ, QR, RS, SP subtend 90◦at Z. Therefore by Lemma 2 the points A′, B′, C′, D′ lie on a circle whose center is the midpoint Y of IZ. Since this circle is the image of the circle ABCD under the considered inver- sion (centered at I), it follows that I, O, Y are collinear, and hence so are I, O, Z. Remark. This is the famous Newton’s theorem for bicentric quadrilaterals.