IMO 2004 SL G6

Origin: GBR | Category: Geometry

Problem

Let P be a convex polygon. Prove that there is a convex hexagon that is contained in P and that occupies at least 75 percent of the area of P.

Solution

Let ABC be the triangle of maximum area S contained in P (it exists because of compactness of P). Draw parallels to BC, CA, AB through A, B, C, respectively, and denote the triangle thus obtained by A1B1C1 (A \inB1C1, etc.). Since each triangle with vertices in P has area at most S, the entire polygon P is contained in A1B1C1. Next, draw lines of support of P parallel to BC, CA, AB and not intersect- ing the triangle ABC. They determine a convex hexagon UaVaUbVbUcVc containing P, with Vb, Uc \inB1C1, Vc, Ua \inC1A1, Va, Ub \inA1B1. Each of the line segments UaVa, UbVb, UcVc contains points of P. Choose such points A0, B0, C0 on UaVa, UbVb, UcVc, respectively. The convex hexagon AC0BA0CB0 is contained in P, because the latter is convex. We prove that AC0BA0CB0 has area at least 3/4 the area of P. Let x, y, z denote the areas of triangles UaBC, UbCA, and UcAB. Then S1 = SAC0BA0CB0 = S + x + y + z. On the other hand, the triangle A1UaVa is similar to \triangleA1BC with similitude \tau = (S −x)/S, and hence its area is \tau 2S = (S −x)2/S. Thus the area of quadrilateral UaVaCB is S −(S −x)2/S = 2z −z2/S. Analogous formulas hold for quadrilaterals UbVbAC and UcVcBA. Therefore SP \leqSUaVaUbVbUcVc = S + SUaVaCB + SUbVbAC + SUcVcBA = S + 2(x + y + z) −x2 + y2 + z2 S \leqS + 2(x + y + z) −(x + y + z)2 3S . Now 4S1−3SP \geq= S−2(x+y+z)+(x+y+z)2/S = (S−x−y−z)2/S \geq0; i.e., S1 \geq3SP/4, as claimed.