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Problem 1553

AIME late

Algebra Difficulty 5.7 Prove it February · United States · 2017

Let $a$ and $b$ be complex numbers satisfying the two equations
$\begin{aligned} a^{3}-3 a b^{2} & =36 \\ b^{3}-3 b a^{2} & =28 i \end{aligned}$
Let $M$ be the maximum possible magnitude of $a$ . Find all $a$ such that $|a|=M$ .

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

Solution:
Notice that
$\begin{aligned} (a-b i)^{3} & =a^{3}-3 a^{2} b i-3 a b^{2}+b^{3} i \\ & =\left(a^{3}-3 a b^{2}\right)+\left(b^{3}-3 b a^{2}\right) i \\ & =36+i(28 i) \\ & =8 \end{aligned}$
so that $a-b i=2+i$ . Additionally
$\begin{aligned} (a+b i)^{3} & =a^{3}+3 a^{2} b i-3 a b^{2}-b^{3} i \\ & =\left(a^{3}-3 a b^{2}\right)-\left(b^{3}-3 b a^{2}\right) i \\ & =36-i(28 i) \\ & =64 \end{aligned}$
It follows that $a-b i=2 \omega$ and $a+b i=4 \omega^{\prime}$ where $\omega, \omega^{\prime}$ are third roots of unity. So $a=\omega+2 \omega^{\prime}$ . From the triangle inequality $|a| \leq|\omega|+\left|2 \omega^{\prime}\right|=3$ , with equality when $\omega$ and $\omega^{\prime}$ point in the same direction (and thus $\omega=\omega^{\prime}$ ). It follows that $a=3,3 \omega, 3 \omega^{2}$ , and so
$a=3,-\frac{3}{2}+\frac{3 i \sqrt{3}}{2},-\frac{3}{2}-\frac{3 i \sqrt{3}}{2}$

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