IMO 1985 SL 5

Origin: ROM

Problem

Let D be the interior of the circle C and let A \inC. Show that the function f : D \toR, f(M) = |MA| |MM′|, where M ′ = (AM \capC, is strictly convex; i.e., f(P) < f(M1)+f(M2) , \forallM1, M2 \inD, M1 ̸= M2, where P is the midpoint of the segment M1M2.

Solution

Let w.l.o.g. circle C have unit radius. For each m \inR, the locus of points M such that f(M) = m is the circle Cm with radius rm = m/(m + 1), that is tangent to C at A. Let Om be the center of Cm. We have to show that if M \inCm and N \inCn, where m, n > 0, then the midpoint P of MN lies inside the circle C(m+n)/2. This is trivial if m = n, so let m ̸= n. For fixed M, P is in the image C′ n of Cn under the homothety with center M and coefficient 1/2. The center of the circle C′ n is at the midpoint of OnM. If we let both M and N vary, P will be on the union of circles with radius rn/2 and centers in the image of Cm under the homothety with center On and coefficient 1/2. Hence P is not outside the circle centered at the midpoint OmOn and with radius (rm + rn)/2. It remains to show that r(m+n)/2 > (rm + rn)/2. But this inequality is easily reduced to (m −n)2 > 0, which is true.