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Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.0 AIME Prove it United States

Problem:
If the system of equations
$\begin{aligned} & |x+y|=99 \\ & |x-y|=c \end{aligned}$
has exactly two real solutions $(x, y)$ , find the value of $c$ .

Solution

Solution:
If $c<0$ , there are no solutions. If $c>0$ then we have four possible systems of linear equations given by $x+y= \pm 99$ , $x-y= \pm c$ , giving four solutions $(x, y)$ . So we must have $c=0$ , and then we do get two solutions ( $x=y$ , so they must both equal $\pm 99 / 2$ ).

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