HMMT February

Algebra Difficulty 4.7 AIME Prove it United States

Problem:

Let $x$ and $y$ be complex numbers such that $x+y=\sqrt{20}$ and $x^{2}+y^{2}=15$ . Compute $|x-y|$ .

Solution

Solution:

We have $(x-y)^{2}+(x+y)^{2}=2\left(x^{2}+y^{2}\right)$ , so $(x-y)^{2}=10$ , hence $|x-y|=\sqrt{10}$ .

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