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Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 10"

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Let's take a closer look at the given condition. We have already changed it into <math>a^3 = -c^3</math>, <math>a \neq -c</math>. Let <math>a = \text{cis}\  
 
Let's take a closer look at the given condition. We have already changed it into <math>a^3 = -c^3</math>, <math>a \neq -c</math>. Let <math>a = \text{cis}\  
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(\theta_a)</math> and <math>c =</math> cis<math>(\theta_c)</math>. By methods such as [[De Moivre's Theorem]], we determine the condition is true if and only if  
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(\theta_a)</math> and <math>c = \text{cis}\ (\theta_c)</math>. By methods such as [[De Moivre's Theorem]], we determine the condition is true if and only if  
 
<cmath> \theta_c = \pm \frac{\pi}{3} - \theta_a </cmath>
 
<cmath> \theta_c = \pm \frac{\pi}{3} - \theta_a </cmath>
 
Since this relationship is supposedly enough to fix <math>k^4</math>, we can set <math>\theta_a = 0 \Rightarrow a = 1</math> without loss of generality.  
 
Since this relationship is supposedly enough to fix <math>k^4</math>, we can set <math>\theta_a = 0 \Rightarrow a = 1</math> without loss of generality.  

Revision as of 14:08, 9 August 2021

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

Our 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 $a = \text{cis}\ (\theta_a)$ and $c = \text{cis}\ (\theta_c)$. 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 is supposedly enough to fix $k^4$, 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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