HMMT-Feb 2019 guts, Problem 19

Complex numbers $a, b, c$ form an equilateral triangle with side length 18 in the complex plane. If $|a+b+c|=36$, find $|b c+c a+a b|$.

Solution: Using basic properties of vectors, we see that the complex number $d=\frac{a+b+c}{3}$ is the center of the triangle. From the given, $|a+b+c|=36 \Longrightarrow|d|=12$. Then, let $a^{\prime}=a-d, b^{\prime}=b-d$, and $c^{\prime}=c-d$. Due to symmetry, $\left|a^{\prime}+b^{\prime}+c^{\prime}\right|=0$ and $\left|b^{\prime} c^{\prime}+c^{\prime} a^{\prime}+a^{\prime} b^{\prime}\right|=0$. Finally, we compute \[ \begin{aligned} |b c+c a+a b| & =\left|\left(b^{\prime}+d\right)\left(c^{\prime}+d\right)+\left(c^{\prime}+d\right)\left(a^{\prime}+d\right)+\left(a^{\prime}+d\right)\left(b^{\prime}+d\right)\right| \\ & =\left|b^{\prime} c^{\prime}+c^{\prime} a^{\prime}+a^{\prime} b^{\prime}+2 d\left(a^{\prime}+b^{\prime}+c^{\prime}\right)+3 d^{2}\right| \\ & =\left|3 d^{2}\right|=3 \cdot 12^{2}=432 \end{aligned} \] $\fbox{432}$.

Answer: 432