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

(Created page with "== Problem 10 == If <math>a,b,c</math> are complex numbers such that <cmath> \begin{eqnarray*} \frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}&=&0,\\ \text{and} \qquad \frac{a\ove...")
 
m (→‎Solution)
 
(4 intermediate revisions by the same user not shown)
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== Solution ==
 
== 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 <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'''
 +
 
Observe that replacing <math>a</math>, <math>b</math>, <math>c</math> with <math>a + z</math>, <math>b + z</math>, <math>c + z</math>, respectively, has no effect on the condition. Then, by setting <math>z = -b</math>, we can set <math>b = 0</math> without loss of generality.  
 
Observe that replacing <math>a</math>, <math>b</math>, <math>c</math> with <math>a + z</math>, <math>b + z</math>, <math>c + z</math>, respectively, has no effect on the condition. Then, by setting <math>z = -b</math>, we can set <math>b = 0</math> without loss of generality.  
 
Substituting this into the condition and clearing denominators yields
 
Substituting this into the condition and clearing denominators yields
 
<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'''
−
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 = \text{cis}
 +
(\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 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>

Latest revision as of 13:08, 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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