36th IMO 1995 shortlist Problem A4 a, b, c are fixed positive reals. Find all positive real solutions x, y, z to: x + y + z = a + b + c and 4xyz - (a 2 x + b 2 y + c 2 z) = abc. Solution
: the unique solution is x = (b+c)/2, y = (c+a)/2, z = (a+b)/2. Dividing by xyz, the second equation becomes 4 = a 2 /yz + b 2 /zx + c 2 /xy + abc/xyz. Put X = a/(yz) 1/2 , Y = b/(zx) 1/2 , Z = c/(xy) 1/2 , then this becomes 4 = X 2 + Y 2 + Z 2 + XYZ (*). All of a, b, c, x, y, z are positive, so X, Y, Z are positive. Hence X 2 < 4, so X < 2. Similarly, Y and Z < 2. If we regard (*) as a quadratic in Z, then we can solve it to get Z = ( (X 2 Y 2 + 16 - 4X 2 - 4Y 2 ) 1/2 - XY)/2 (the other solution is negative and hence ruled out). The key is to subsitute X = 2 sin u, Y = 2 sin v, then Z = 2(cos u cos v - sin u sin v) = 2 cos(u + v). So we have a = 2 (yz) 1/2 sin u, b = 2 (zx) 1/2 sin v, c = 2 (xy) 1/2 (cos u cos v - sin u sin v). So far we have not used the first equation. If we require that it holds then we have x + y + z = 2 (yz) 1/2 sin u + 2 (zx) 1/2 sin v + 2 (xy) 1/2 (cos u cos v - sin u sin v) (**). Writing x = ( x 1/2 sin v) 2 + (x 1/2 cos v) 2 , y = (y 1/2 sin u) 2 + (y 1/2 cos u) 2 , z = (z 1/2 ) 2 , we can write (**) as (x 1/2 cos v - y 1/2 cos u) 2 + (x 1/2 sin v + y 1/2 sin u - z 1/2 ) 2 = 0. Squares are non-negative, so each term must be zero. Hence z 1/2 = x 1/2 sin v + y 1/2 sin u = 1/z 1/2 (b/2 + a/2). Hence z = (a + b)/2. But the problem is symmetrical. So we must also have x = (b + c)/2, y = (c + a)/2. It is not hard to check that this is indeed a solution. We get 8xyz - 2(a 2 x + b 2 y + c 2 z) = (b + c)(c + a)(a + b) - (a 2 (b+c) + b 2 (c+a) + c 2 (a+b) ) = 2abc (after some cancellation). 36th IMO shortlist 1995 © John Scholes jscholes@kalva.demon.co.uk 4 Sep 2002