USAMO GuideUSAMOGuide
Back to Problem
1900
Symmetric System — Editorial
algebrasystems of equationssymmetric expressions
Let S1=x2+y2+z2S_1 = x^2+y^2+z^2 and S2=xy+yz+zx.S_2 = xy+yz+zx. Adding the three equations:
2S1+S2=75+16+39=130.(1)2S_1 + S_2 = 75 + 16 + 39 = 130. \tag{1}
Subtracting pairs gives (x−z)(x+y+z)=59,(x-z)(x+y+z) = 59, (x−y)(x+y+z)=23,(x-y)(x+y+z) = 23, (y−z)(x+y+z)=36.(y-z)(x+y+z) = 36. Squaring and summing:
(x+y+z)2[(x−z)2+(x−y)2+(y−z)2]=592+232+362=5306.(x+y+z)^2\bigl[(x-z)^2+(x-y)^2+(y-z)^2\bigr] = 59^2+23^2+36^2 = 5306.
Using (x−z)2+(x−y)2+(y−z)2=2(S1−S2)(x-z)^2+(x-y)^2+(y-z)^2 = 2(S_1-S_2) and (x+y+z)2=S1+2S2,(x+y+z)^2 = S_1+2S_2, with S1=(130−S2)/2S_1=(130-S_2)/2 from (1):
5306=(130+3S2)(130−3S2)2,5306 = \frac{(130+3S_2)(130-3S_2)}{2},
giving 9S22=6288,9S_2^2 = 6288, so 3S22=2096.3S_2^2 = 2096. Therefore 3(xy+yz+zx)2=2096.3(xy+yz+zx)^2 = 2096.