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

m (→‎Solution)
m (→‎Solution)
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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 <math>k^4</math>. Counterintuitively, this very fact offers lots of information.
 
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 <math>k^4</math>. Counterintuitively, this very fact offers lots of information.
 +
  
 
'''Degree of Freedom 1: Translation'''
 
'''Degree of Freedom 1: Translation'''
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<cmath> a^2 - ac + c^2 = \frac{a^3 + c^3}{a + c} = 0. </cmath>
 
<cmath> a^2 - ac + c^2 = \frac{a^3 + c^3}{a + c} = 0. </cmath>
 
Then <math>a^3 = -c^3</math>, with <math>a \neq -c</math>; this implies <math>|a| = |c|</math>.
 
Then <math>a^3 = -c^3</math>, with <math>a \neq -c</math>; this implies <math>|a| = |c|</math>.
 +
  
 
'''Degree of Freedom 2: Dilation'''
 
'''Degree of Freedom 2: Dilation'''
  
 
Observe that replacing <math>a</math>, <math>b</math>, <math>c</math>, with <math>ra</math>, <math>rb</math>, <math>rc</math>, respectively, has no effect on the condition. Then, an appropriate <math>r</math> can be chosen such that <math>|a| = |c| = 1</math>; that is, without loss of generality, <math>|a| = |c| = 1</math>.  
 
Observe that replacing <math>a</math>, <math>b</math>, <math>c</math>, with <math>ra</math>, <math>rb</math>, <math>rc</math>, respectively, has no effect on the condition. Then, an appropriate <math>r</math> can be chosen such that <math>|a| = |c| = 1</math>; that is, without loss of generality, <math>|a| = |c| = 1</math>.  
 +
  
 
'''Degree of Freedom 3: Rotation'''
 
'''Degree of Freedom 3: Rotation'''
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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 =</math> cis<math>(\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  
 
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 =</math> cis<math>(\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  
 
<cmath> \theta_c = \pm \frac{\pi}{3} - \theta_a </cmath>
 
<cmath> \theta_c = \pm \frac{\pi}{3} - \theta_a </cmath>
−
Since this relationship supposedly fixes <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.  
  
 
From here, we determine <math>\theta_c = \pm \frac{\pi}{3}</math> and <math>c = \frac{1}{2} \pm \frac{\sqrt{3}}{2}</math>. Then we can compute  
 
From here, we determine <math>\theta_c = \pm \frac{\pi}{3}</math> and <math>c = \frac{1}{2} \pm \frac{\sqrt{3}}{2}</math>. Then we can compute  
 
<cmath> k = \pm \sqrt{3}i \Rightarrow k^4 = \boxed{009}. </cmath>
 
<cmath> k = \pm \sqrt{3}i \Rightarrow k^4 = \boxed{009}. </cmath>

Revision as of 13:04, 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 $a =$ 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 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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