2022 AMC 12B Problem 24

Attempt Problem 24 of the 2022 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

The figure below depicts a regular 77-gon inscribed in a unit circle.
What is the sum of the 44th powers of the lengths of all 2121 of its edges and diagonals?
4949
9898
147147
168168
196196
Answer: C
Concepts:chordroots of unitytrigonometric identity
Small hint:
A chord between vertices dd apart has squared length 2−2cos⁡2πd7;2 - 2\cos\dfrac{2\pi d}{7}; there are 77 of each type d=1,2,3d = 1, 2, 3
Big hint:
Use ∑d=13cos⁡2πd7=−12\sum_{d=1}^{3} \cos\dfrac{2\pi d}{7} = -\dfrac12 to evaluate ∑(2−2cos⁡2πd7)2\sum \left(2 - 2\cos\dfrac{2\pi d}{7}\right)^2
Solution:
A chord joining two vertices dd steps apart has squared length 2−2cos⁡2πd7,2 - 2\cos\dfrac{2\pi d}{7}, and there are 77 chords for each of d=1,2,3.d = 1, 2, 3. The required sum is 7∑d=13(2−2cos⁡2πd7)2. 7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2.
Using ∑d=13cos⁡2πd7=−12\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12 and ∑d=13cos⁡22πd7=54,\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54, the inner sum expands to 4(3+1+54)=21.4\left(3 + 1 + \tfrac54\right) = 21.
Therefore the total is 7⋅21=147.7 \cdot 21 = 147.
Thus, the correct answer is C.
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