Northeastern WOOTers Mock AIME I Problems/Problem 10

Problem 10

If $a,b,c$ are complex numbers such that $\begin{eqnarray*} \frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}&=&0,\\ \text{and} \qquad \frac{a\overline{b}+b\overline{c}+c\overline{a}-\overline{a}b-\overline{b}c-\overline{c}a}{\left(a-b\right)\left(\overline{a-b}\right)}&=&k, \end{eqnarray*}$ then find the value of $k^4$ .

Solution

The strategy is to take advantage of degrees of freedom. The given condition appears extremely weak (that is, it offers little information), yet apparently it uniquely determines $k^4$ . Counterintuitively, this very fact offers lots of information.

Degree of Freedom 1: Translation Observe that replacing $a$ , $b$ , $c$ with $a + z$ , $b + z$ , $c + z$ , respectively, has no effect on the condition. Then, by setting $z = -b$ , we can set $b = 0$ without loss of generality. Substituting this into the condition and clearing denominators yields $a^2 - ac + c^2 = \frac{a^3 + c^3}{a + c} = 0.$ Then $a^3 = -c^3$ , with $a \neq -c$ ; this implies $|a| = |c|$ .

Degree of Freedom 2: Dilation Observe that replacing $a$ , $b$ , $c$ , with $ra$ , $rb$ , $rc$ , respectively, has no effect on the condition. Then, an appropriate $r$ can be chosen such that $|a| = |c| = 1$ ; that is, without loss of generality, $|a| = |c| = 1$ .

Degree of Freedom 3: Rotation Let's take a closer look at the given condition. We have already changed it into $a^3 = -c^3$ , $a \neq -c$ . Let $a =$ cis $(\theta_a)$ and $c =$ cis $(\theta_c)$ . By methods such as De Moivre's Theorem, we determine the condition is true if and only if $\theta_c = \pm \frac{\pi}{3} - \theta_a$ Since this relationship supposedly fixes $k^4$ , we can set $\theta_a = 0 \Rightarrow a = 1$ without loss of generality.

From here, we determine $\theta_c = \pm \frac{\pi}{3}$ and $c = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$ . Then we can compute $k = \pm \sqrt{3}i \Rightarrow k^4 = \boxed{009}.$

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Northeastern WOOTers Mock AIME I Problems/Problem 10

Problem 10

Solution