Difference between revisions of "2021 AMC 12A Problems/Problem 22"
m (→Solution) |
m (→Solution) |
||
Line 39: | Line 39: | ||
<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> | <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> | ||
− | We know that <math>\cos \alpha \cos \beta = | + | We know that <math>\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}</math> |
By plugging all the parts in we get: | By plugging all the parts in we get: |
Revision as of 13:57, 12 February 2021
Contents
Problem
Suppose that the roots of the polynomial are and , where angles are in radians. What is ?
Solution
Part 1: solving for c
Notice that
is the negation of the product of roots by Vieta's formulas
Multiply by
Then use sine addition formula backwards:
Part 2: starting to solve for b
is the sum of roots two at a time by Vieta's
We know that
By plugging all the parts in we get:
Which ends up being:
Which is shown in the next part to equal , so
Part 3: solving for a and b as the sum of roots
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 .
~Tucker
Video Solution by OmegaLearn (Euler's Identity + Vieta's )
~ pi_is_3.14
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.