Track / Stage 5 / 125 of 400 #1365 of 2604

Problem 1365

AIME late

Algebra Difficulty 5.2 Prove it Spring Mathematical Tournament · Bulgaria

Find all real numbers $a$ and $b$ such that the system
$\begin{array}{llc} \text{system} & \left\{ \begin{array}{l} x + a = y + b \\ x^2 - a = 2y \end{array} \right. & \text{has unique solution } (x_0, y_0) \text{ and it satisfies the equality} \\ & x_0^2 + y_0^2 = 1025. \end{array}$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The given system is equivalent to
$\left| \begin{array}{l} x^2 - 2x + 2b - 3a = 0 \\ x + a = y + b \end{array} \right.$
It has a unique solution $(x_0, y_0)$ if the quadratic equation $x^2 - 2x + 2b - 3a = 0$ has a unique root $x_0$ . This means that $D = 1 - 2b + 3a = 0$ and $x_0 = 1$ . The condition $x_0^2 + y_0^2 = 1025$ gives $y_0^2 = 1024$ , i.e. $y_0 = \pm 2$ . For $y_0 = -2$ we get $a = x_0^2 - 2y_0 = 5$ and $b = x_0 + a - y_0 = 8$ , and for $y_0 = 2$ we get $a = x_0^2 - 2y_0 = -3$ and $b = x_0 + a - y_0 = -4$ .

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