IMO 1995 SL A4

Let a, b, and c be given positive real numbers. Determine all

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IMO 1995 SL A4

Origin: USA | Category: Algebra

Problem

Let a, b, and c be given positive real numbers. Determine all positive real numbers x, y, and z such that x + y + z = a + b + c and 4xyz −(a2x + b2y + c2z) = abc.

Solution

The second equation is equivalent to a2 yz + b2 zx + c2 xy + abc xyz = 4. Let x1 = a \sqrtyz, y1 = b \sqrtzx, z1 = c \sqrtxy. Then x2 1 + y2 1 + z2 1 + x1y1z1 = 4, where 0 < x1, y1, z1 < 2. Regarding this as a quadratic equation in z1, the discriminant (4−x2 1)(4−y2

  1. suggests that we let x1 = 2 sin u, y1 = 2 sin v, 0 < u, v < \pi/2. Then it is directly shown that z1 will be exactly 2 cos(u+v) as the only positive solution of the quadratic equation. Thus a = 2\sqrtyz sin u, b = 2\sqrtxz sin v, c = 2\sqrtxy(cos u cos v −sin u sin v), so from x + y + z −a −b −c = 0 we obtain (\sqrtx cos v −\sqrty cos u)2 + (\sqrtx sin v + \sqrty sin u −\sqrtz)2 = 0,

which implies \sqrtz = \sqrtx sin v + \sqrty sin u = 1 2(y1 \sqrtx + x1 \sqrty) = 1  b \sqrtzx \sqrtx + a \sqrtyz \sqrty  . Therefore z = a+b 2 . Similarly, x = b+c and y = c+a 2 . It is clear that the triple (x, y, z) =  b+c 2 , c+a 2 , a+b  is indeed a (unique) solution of the given system of equations. Second solution. Put x = b+c −u, y = c+a −v, z = a+b −w, where u \leq b+c 2 , v \leq c+a 2 , w \leq a+b and u + v + w = 0. The equality abc + a2x + b2y + c2z = 4xyz becomes 2(au2 + bv2 + cw2 + 2uvw) = 0. Now uvw > 0 is clearly impossible. On the other hand, if uvw \leq0, then two of u, v, w are nonnegative, say u, v \geq0. Taking into account w = −u −v, the above equality reduces to 2[(a+c−2v)u2 +(b+c−2u)v2 +2cuv] = 0, so u = v = 0. Third solution. The fact that we are given two equations and three vari- ables suggests that this is essentially a problem on inequalities. Setting f(x, y, z) = 4xyz −a2x −b2y −c2z, we should show that max f(x, y, z) = abc, for 0 < x, y, z, x + y + z = a + b + c, and find when this value is attained. Thus we apply Lagrange multipliers to F(x, y, z) = f(x, y, z) − \lambda(x + y + z −a −b −c), and obtain that f takes a maximum at (x, y, z) such that 4yz −a2 = 4zx −b2 = 4xy −c2 = \lambda and x + y + z = a + b + c. The only solution of this system is (x, y, z) =  b+c 2 , c+a 2 , a+b  .