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...") |
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== Solution == | == 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 <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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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 | ||
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'''Degree of Freedom 2: Dilation''' | '''Degree of Freedom 2: Dilation''' | ||
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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 =</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> |
Revision as of 13:02, 9 August 2021
Problem 10
If are complex numbers such that then find the value of .
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 . Counterintuitively, this very fact offers lots of information.
Degree of Freedom 1: Translation
Observe that replacing , , with , , , respectively, has no effect on the condition. Then, by setting , we can set without loss of generality. Substituting this into the condition and clearing denominators yields Then , with ; this implies .
Degree of Freedom 2: Dilation
Observe that replacing , , , with , , , respectively, has no effect on the condition. Then, an appropriate can be chosen such that ; that is, without loss of generality, .
Degree of Freedom 3: Rotation
Let's take a closer look at the given condition. We have already changed it into , . Let cis and cis. By methods such as De Moivre's Theorem, we determine the condition is true if and only if Since this relationship supposedly fixes , we can set without loss of generality.
From here, we determine and . Then we can compute