Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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<math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math> | <math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math> | ||
− | Therefore, we have <math>a = -\left(-\frac{1}2\right) = \frac{1}2 | + | Therefore, we have <math>a = -\left(-\frac{1}2\right) = \frac{1}2</math> |
Finally multiply <math>abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}32</math> or D. | Finally multiply <math>abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}32</math> or D. |
Revision as of 17:48, 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
This comes out to , but I forget how.
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
Finally multiply or D.
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.