HMMT February

Algebra Difficulty 5.5 AIME, harder Find the answer

Suppose that $x, y$ , and $z$ are complex numbers of equal magnitude that satisfy $x+y+z=-\frac{\sqrt{3}}{2}-i \sqrt{5}$ and $x y z=\sqrt{3}+i \sqrt{5}.$ If $x=x_{1}+i x_{2}, y=y_{1}+i y_{2}$ , and $z=z_{1}+i z_{2}$ for real $x_{1}, x_{2}, y_{1}, y_{2}, z_{1}$ , and $z_{2}$ , then $\left(x_{1} x_{2}+y_{1} y_{2}+z_{1} z_{2}\right)^{2}$ can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100 a+b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

From the conditions, it is clear that $a, b, c$ all have magnitude $\sqrt{2}$ . Conjugating the first equation gives $2\left(\frac{a b+b c+c a}{a b c}\right)=-\frac{\sqrt{3}}{2}+i \sqrt{5}$ , which means $a b+b c+c a=\left(-\frac{\sqrt{3}}{4}+i \frac{\sqrt{5}}{2}\right)(\sqrt{3}+i \sqrt{5})=\frac{-13+i \sqrt{15}}{4}$ . Then, $\begin{aligned} a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} & =\frac{1}{2} \operatorname{Im}\left(a^{2}+b^{2}+c^{2}\right) \\ & =\frac{1}{2} \operatorname{Im}\left((a+b+c)^{2}\right)-\operatorname{Im}(a b+b c+c a) \\ & =\frac{\sqrt{15}}{4} \end{aligned}$ so the answer is 1516.

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