IMO 2001 SL G6

Origin: IND | Category: Geometry

Problem

Let ABC be a triangle and P an exterior point in the plane of the triangle. Suppose AP, BP, CP meet the sides BC, CA, AB (or ex- tensions thereof) in D, E, F, respectively. Suppose further that the areas of triangles PBD, PCE, PAF are all equal. Prove that each of these areas is equal to the area of triangle ABC itself.

Solution

By Ceva’s theorem, we can choose real numbers x, y, z such that −−\to BD −−\to DC = z y , −−\to CE −\to EA = x z , and −\to AF −−\to FB = y x. The point P lies outside the triangle ABC if and only if x, y, z are not all of the same sign. In what follows, SX will denote the signed area of a figure X. Let us assume that the area SABC of \triangleABC is 1. Since SPBC : SPCA : SPAB = x : y : z and SPBD : SPDC = z : y, it follows that SPBD = z y+z x x+y+z. Hence SPBD = y(y+z) xyz x+y+z, SPCE = z(z+x) xyz x+y+z, SPAF = x(x+y) xyz x+y+z. By the condition of the problem we have |SPBD| = |SPCE| = |SPAF |, or |x(x + y)| = |y(y + z)| = |z(z + x)|. Obviously x, y, z are nonzero, so that we can put w.l.o.g. z = 1. At least two of the numbers x(x+y), y(y +1), 1(1+x) are equal, so we can assume that x(x + y) = y(y + 1). We distinguish two cases: (i) x(x + y) = y(y + 1) = 1 + x. Then x = y2 + y −1, from which we obtain (y2 + y −1)(y2 + 2y −1) = y(y + 1). Simplification gives y4 + 3y3 −y2 −4y + 1 = 0, or (y −1)(y3 + 4y2 + 3y −1) = 0. If y = 1, then also z = x = 1, so P is the centroid of \triangleABC, which is not an exterior point. Hence y3 + 4y2 + 3y −1 = 0. Now the signed area of each of the triangles PBD, PCE, PAF equals SPAF = yz (x + y)(x + y + z)

y (y2 + 2y −1)(y2 + 2y) = y3 + 4y2 + 3y −2 = −1. It is easy to check that not both of x, y are positive, implying that P is indeed outside \triangleABC. This is the desired result. (ii) x(x + y) = y(y + 1) = −1 −x. In this case we are led to f(y) = y4 + 3y3 + y2 −2y + 1 = 0. We claim that this equation has no real solutions. In fact, assume that y0 is a real root of f(y). We must have y0 < 0, and hence u = −y0 > 0 satisfies 3u3 −u4 = (u + 1)2. On the other hand,

3u3 −u4 = u3(3 −u) = 4u u u

(3 −u) \leq4u u/2 + u/2 + 3 −u 3 = 4u \leq(u + 1)2, where at least one of the inequalities is strict, a contradiction. Remark. The official solution was incomplete, missing the case (ii).