2022 AMC 12B Problems/Problem 24
Contents
- 1 Problem
- 2 Solution 1 (Complex Numbers)
- 3 Solution 2 (Trigonometry)
- 4 Solution 3 (Complex Numbers and Trigonometry)
- 5 Solution 4 (Trigonometry)
- 6 Solution 5 (Law of Cosines)
- 7 Solution 6 (Ruler Measure)
- 8 Solution 7 (Pythagorean Theorem and Trig)
- 9 Solution 8 (Roots of Unity)
- 10 Solution 9 (Inductive Reasoning)
- 11 Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)
- 12 Video Solution
- 13 See Also
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
For explanation see supplement.
~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.
For explanation see supplement.
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 4 (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 , , and 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 of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by .
(Use the identity that .)
For explanation of this identity see supplement.
- 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 have successfully finished the problem.
~sirswagger21
Solution 7 (Pythagorean Theorem and Trig)
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle which are for integer Then, we notice there are three types of diagonals: the ones with chords of arcs and We notive there are here are of each type of diagonal. Then, we use the pythagorean theorem to find the distance from to : By the cosine double angle identity, This means that Substituting this in, Summing this up for (These equalities are based on and ) (For explanation of this see supplement.) Finally, because there are of each type of diagonal, the answer is
Solution 8 (Roots of Unity)
Place the figure on the complex plane and let (a th root of unity). The vertices of the -gon are . We wish to find The second expression is more convenient to work with. The factor of is because it double-counts each edge (and includes when , but the length is so it doesn't matter).
Recall the identity . Since , , and
But notice that, for a fixed value of , over all values of from to , by properties of modular arithmetic with (or expanding), each of the -terms takes on every value in exactly once. Since ,
Fun fact: This generalizes to any regular -gon with to obtain though for even , and take on every value in exactly twice. These are the th roots of unity, so they still sum to .
This also explains why doesn't follow the pattern. For (one edge of length ), since and similarly for , the constant term in the summation becomes instead of . For (no edges), the constant term becomes .
-maxamc
~edited by emerald_block
Solution 9 (Inductive Reasoning)
This is how I solve this problem:
It's easy to solve for -gon, -gon, and -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For -gon, the sum is equal to times the th power of an edge. Thus,
For -gon, the sum is equal to times the th power of an edge, and times the th power of the diagonal. Thus,
For -gon, the sum is equal to times the th power of an edge, times the th power of the short diagonal, and times the th power of the long diagonal. Thus,
Then, I quickly noticed that , , and . So reasonably, it will work out this formula, . (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
By inductive reasoning, we got .
- Prof. Joker
Supplement (Explanation of why cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2)
The sum of all roots of unity is , therefore, the sum of the real parts of the roots of unity is .
Consider the roots of unity of where is odd on the unit circle, the cosine (real parts) of all the roots under the x-axis is the same as the cosine of all the roots above the x-axis because of symmetry. Hence, times the sum of the cosines of all the roots above the x-axis plus is .
In the context of this problem, . Hence,
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 |
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