Difference between revisions of "2022 AMC 12B Problems/Problem 24"
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== Solution 5 (Trigonometry) == | == Solution 5 (Trigonometry) == | ||
− | This solution follows the same steps as the trigonometry solutions (Solutions 2 | + | This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers: |
<cmath>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</cmath> | <cmath>\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}</cmath> | ||
Line 180: | Line 180: | ||
We end up with <cmath>2S^2 - 5S - 3 = 0.</cmath> | We end up with <cmath>2S^2 - 5S - 3 = 0.</cmath> | ||
Using the quadratic formula, we find the solutions for <math>S</math> to be <math>-\frac{1}{2}</math> and <math>3</math>. Because <math>3</math> is impossible, <math>S = -\frac{1}{2}</math>. | Using the quadratic formula, we find the solutions for <math>S</math> to be <math>-\frac{1}{2}</math> and <math>3</math>. Because <math>3</math> is impossible, <math>S = -\frac{1}{2}</math>. | ||
− | With this result, following similar to steps to Solutions 2 | + | With this result, following similar to steps to Solutions 2 and 3 will get <math>\boxed{\textbf{(C)} \ 147}</math> |
~lordf | ~lordf |
Revision as of 12:48, 10 January 2023
Contents
Problem
The figure below depicts a regular -gon inscribed in a unit circle. What is the sum of the th powers of the lengths of all of its edges and diagonals?
Solution 1 (Complex Numbers)
There are segments whose lengths are , segments whose lengths are , segments whose lengths are .
Therefore, the sum of the th powers of these lengths is where the fourth from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Trigonometry)
There are segments whose lengths are , segments whose lengths are , segments whose lengths are .
Therefore, the sum of the th powers of these lengths is where the second from the last equality follows from the property that
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Complex Numbers and Trigonometry)
As explained in Solutions 1 and 2, what we are trying to find is . Using trig we get Like in the second solution, we also use the fact that , which admittedly might need some explanation. Notice that In the brackets we have the sum of the roots of the polynomial . These sum to by Vieta’s formulas, and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question: ~obscene_kangaroo
Solution 5 (Trigonometry)
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
We end up with Using the quadratic formula, we find the solutions for to be and . Because is impossible, . With this result, following similar to steps to Solutions 2 and 3 will get
~lordf
Solution 5 (Law of Cosines)
Let x,y,z be the lengths of the chords with arcs , and respectively.
Then by the law of cosines we get:
The answer is then just (since there's 7 of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by 7.
(Uses the identity that )
- SAHANWIJETUNGA
Solution 6 (Ruler Measure)
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by .
Measuring the sides of the circle gets cm. The shorter diagonals are cm, and the longest diagonals measure cm. Thus, we'd like to estimate
We know is slightly less than Let's approximate it as 1 for now. Thus,
Next, is slightly more than We know slightly more than so we can approximate as Thus,
Finally, is slightly less than We say it's around so then
Adding what we have, we get as our estimate. We see is very close to our estimate, so we circle it and are happy that we successfully cheesed an AMC 12B problem 24.
~sirswagger21
Video Solution
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.