IMO 1983 SL 17

Origin: GDR

Problem

Let P1, P2, . . . , Pn be distinct points of the plane, n \geq2. Prove that max 1\leqi<j\leqn PiPj > \sqrt 2 (\sqrtn −1) min 1\leqi<j\leqn PiPj.

Solution

Set a = min PiPj, b = max PiPj. We use the following lemma. Lemma. There exists a disk of radius less than or equal to b/ \sqrt 3 containing all the Pi’s. Assuming that this is proved, the disks with center Pi and radius a/2 are disjoint and included in a disk of radius b/ \sqrt 3 + a/2; hence comparing areas, n\pi \cdot a2 4 < \pi \cdot  b \sqrt

• a/2 2 and b > \sqrt 3/2 \cdot (\sqrtn −1)a. Proof of the lemma. If a nonobtuse triangle with sides a \geqb \geqc has a circumscribed circle of radius R, we have R = a/(2 sin \alpha) \leqa/ \sqrt

Now we show that there exists a disk D of radius R containing A = {P1, . . . , Pn} whose border C is such that C \capA is not included in an open semicircle, and hence contains either two diametrically opposite points and R \leqb/2, or an acute-angled triangle and R \leqb/ \sqrt 3. Among all disks whose borders pass through three points of A and that contain all of A, let D be the one of least radius. Suppose that C \capA is contained in an arc of central angle less than 180◦, and that Pi, Pj are its endpoints. Then there exists a circle through Pi, Pj of smaller radius that contains A, a contradiction. Thus D has the required property, and the assertion follows.