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.

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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
Difficulty rating: 2370
Solution:

A chord joining two vertices dd steps apart has squared length 22cos2πd7,2 - 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 7d=13(22cos2πd7)2. 7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2.

Using d=13cos2πd7=12\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12 and d=13cos22πd7=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 721=147.7 \cdot 21 = 147.

Thus, the correct answer is C.

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