Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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Part 1: solving for <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math> | Part 1: solving for <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math> | ||
+ | |||
+ | <math>c</math> is the negation of the product of roots by Vieta's formulas | ||
<math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math> | <math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math> | ||
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− | This is in progress - I am working on | + | Part 2: in progress - solving for b |
+ | |||
+ | <math>b</math> is the sum of roots two at a time by Vieta's | ||
+ | |||
+ | <math>b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7</math> | ||
+ | |||
+ | |||
+ | Part 3: solving for a | ||
+ | |||
+ | <math>a</math> is the negation of the sum of roots | ||
+ | |||
+ | <math>a = \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7</math> | ||
+ | |||
+ | The real values of the 7th roots of unity are: <math>1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7</math> and they sum to <math>0</math>. | ||
+ | |||
+ | If we subtract 1, and condense identical terms, we get: | ||
+ | |||
+ | <math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math> | ||
+ | |||
+ | Therefore, we have $a = -\frac{1}2 | ||
+ | |||
+ | |||
+ | |||
+ | This is in progress - I am working on the solution for b. | ||
~Tucker | ~Tucker |
Revision as of 17:36, 11 February 2021
Problem
Suppose that the roots of the polynomial are
and
, where angles are in radians. What is
?
Solution
Part 1: solving for
is the negation of the product of roots by Vieta's formulas
Multiply by
Then use sine addition formula backwards:
Part 2: in progress - solving for b
is the sum of roots two at a time by Vieta's
Part 3: solving for a
is the negation of the sum of roots
The real values of the 7th roots of unity are: and they sum to
.
If we subtract 1, and condense identical terms, we get:
Therefore, we have $a = -\frac{1}2
This is in progress - I am working on the solution for b.
~Tucker
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.