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Algebra Difficulty 4.8 AIME Prove it Bulgaria

If x,y,z∈Rx, y, z \in \mathbb{R} satisfy the nonlinear system of equations
∣x−y+z−1=0xy+2z2−6z+1=0, \left| \begin{array}{l} x - y + z - 1 = 0 \\ xy + 2z^2 - 6z + 1 = 0 \end{array} \right. ,

find the maximum of (x−1)2+(y+1)2(x-1)^2 + (y+1)^2.

Solution

Plugging the first equation into the second, we get
x(x+z−1)+2z2−6z+1=x2+(z−1)x+2z2−6z+1=0, x(x+z-1)+2z^2-6z+1=x^2+(z-1)x+2z^2-6z+1=0,
which is quadratic with respect to xx. The discriminant is D=−7z2+22z−3D = -7z^2 + 22z - 3, thus we need z∈[17,3]z \in [\frac{1}{7}, 3]. Note that
(x−1)2+(y+1)2=(y−x)2+2xy+2(y−x)+2=−3z2+12z−1. (x-1)^2 + (y+1)^2 = (y-x)^2 + 2xy + 2(y-x) + 2 = -3z^2 + 12z - 1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.