IMO 2002 SL G5

Origin: AUS | Category: Geometry

Problem

For any set S of five points in the plane, no three of which are collinear, let M(S) and m(S) denote the greatest and smallest areas, respectively, of triangles determined by three points from S. What is the minimum possible value of M(S)/m(S)?

Solution

When S is the set of vertices of a regular pentagon, then it is easily verified that M(S) m(S) = 1+ \sqrt = \alpha. We claim that this is the best possible. Let A, B, C, D, E be five arbitrary points, and assume that \triangleABC has the area M(S). We claim that some triangle has area less than or equal to M(S)/\alpha. Construct a larger triangle A′B′C′ with C \inA′B′ \parallelAB, A \inB′C′ \parallelBC, B \inC′A′ \parallelCA. The point D, as well as E, must lie on the same side of B′C′ as BC, for otherwise \triangleDBC would have greater area than \triangleABC. A similar result holds for the other edges, and therefore D, E lie inside the triangle A′B′C′ or on its boundary. Moreover, at least one of the triangles A′BC, AB′C, ABC′, say ABC′, contains neither D nor E. Hence we can assume that D, E are contained inside the quadrilateral A′B′AB. An affine linear transformation does not change the ratios between areas. Thus if we apply such an affine transformation mapping A, B, C into the vertices ABMCN of a regular pentagon, we won’t change M(S)/m(S). If now D or E lies inside ABMCN, then we are done. Suppose that both D and E are inside the triangles CMA′, CNB′. Then CD, CE \leqCM (because CM = CN = CA′ = CB′) and \angleDCE is either less than or equal to 36◦or greater than or equal to 108◦, from which we obtain that the area of \triangleCDE cannot exceed the area of \triangleCMN = M(S)/\alpha. This completes the proof.