Difference between revisions of "1994 AIME Problems/Problem 13"
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Thus, <math>13-\dfrac{1}{x} = \omega</math> where <math>\omega = e^{i(\pi n/5+\pi/10)}</math> where <math>n</math> is an integer. | Thus, <math>13-\dfrac{1}{x} = \omega</math> where <math>\omega = e^{i(\pi n/5+\pi/10)}</math> where <math>n</math> is an integer. | ||
− | We see that <math>\dfrac{1}{x}=13-\omega</math>. Thus, <cmath>\dfrac{1}{x\overline{x}}=(13-\omega)(13-\overline{\omega})=169-13(\omega+\overline{\omega})+\omega\overline{\omega}=170-13(\omega+\overline{\omega})</cmath> | + | We see that <math>\dfrac{1}{x}=13-\omega</math>. Thus, <cmath>\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})</cmath> |
Summing over all terms: <cmath>\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})</cmath> | Summing over all terms: <cmath>\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})</cmath> |
Latest revision as of 23:46, 6 August 2021
Contents
Problem
The equation
has 10 complex roots where the bar denotes complex conjugation. Find the value of
Solution 1
Let . After multiplying the equation by , .
Using DeMoivre, where is an integer between and .
.
Since , after expanding. Here ranges from 0 to 4 because two angles which sum to are involved in the product.
The expression to find is .
But so the sum is .
Solution 2
Divide both sides by to get
Rearranging:
Thus, where where is an integer.
We see that . Thus,
Summing over all terms:
However, note that from drawing the numbers on the complex plane, our answer is just
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.