IMO 2003 SL A1

Origin: USA | Category: Algebra

Problem

Let aij, i = 1, 2, 3, j = 1, 2, 3, be real numbers such that aij is positive for i = j and negative for i ̸= j. Prove that there exist positive real numbers c1, c2, c3 such that the num- bers a11c1 + a12c2 + a13c3, a21c1 + a22c2 + a23c3, a31c1 + a32c2 + a33c3 are all negative, all positive, or all zero.

Solution

Consider the points O(0, 0, 0), P(a11, a21, a31), Q(a12, a22, a32), R(a13, a23, a33) in three-dimensional Euclidean space. It is enough to find a point U(u1, u2, u3) in the interior of the triangle PQR whose coordinates are all positive, all negative, or all zero (indeed, then we have −−\to OU = c1 −−\to OP + c2 −−\to OQ + c3 −−\to OR for some c1, c2, c3 > 0 with c1 + c2 + c3 = 1). Let P ′(a11, a21, 0), Q′(a12, a22, 0), and R′(a13, a23, 0) be the projections of P, Q, and R onto the Oxy plane. We see that P ′, Q′, R′ lie in the fourth, second, and third quadrants, respectively. We have the following two cases: (i) O is in the exterior of \triangleP ′Q′R′. Set S′ = OR′ \capP ′Q′ and let S be the point of the segment PQ that projects to S′. The point S has its z coordinate negative (be- cause the z coordinates of P and Q are negative). Thus any point y x O R′ Q′ P ′ S′ of the segment SR sufficiently close to S has all coordinates negative. (ii) O is in the interior or on the boundary of \triangleP ′Q′R′. Let T be the point in the plane PQR whose projection is O. If T = O, then all coordinates of T are zero, and we are done. Otherwise O is interior to \triangleP ′Q′R′. Suppose that the z coordinate of T is positive (negative). Since x and y coordinates of T are equal to 0, there is a point U inside PQR close to T with both x and y coordinates positive (respectively negative), and this point U has all coordinates of the same sign.