HMMT February, 2019

Algebra Difficulty 5.9 AIME, harder Prove it United States

Problem:

For any real number $\alpha$ , define
$\operatorname{sign}(\alpha)= \begin{cases}+1 & \text{ if } \alpha>0 \\ 0 & \text{ if } \alpha=0 \\ -1 & \text{ if } \alpha<0\end{cases}$
How many triples $(x, y, z) \in \mathbb{R}^3$ satisfy the following system of equations
$\begin{aligned} & x=2018-2019 \cdot \operatorname{sign}(y+z) \\ & y=2018-2019 \cdot \operatorname{sign}(z+x) \\ & z=2018-2019 \cdot \operatorname{sign}(x+y) ? \end{aligned}$

Solution

Solution:

Since $\operatorname{sign}(x+y)$ can take one of 3 values, $z$ can be one of 3 values: 4037, 2018, or $-1$ . The same is true of $x$ and $y$ . However, this shows that $x+y$ cannot be $0$ , so $z$ can only be 4037 or $-1$ . The same is true of $x$ and $y$ . Now note that, if any two of $x, y, z$ are $-1$ , then the third one must be 4037. Furthermore, if any one of $x, y, z$ is 4037, then the other two must be $-1$ . Thus, the only possibility is to have exactly two of $x, y, z$ be $-1$ and the third one be 4037. This means that the only remaining triples are $(-1,-1,4037)$ and its permutations. These all work, so there are exactly 3 ordered triples.

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