2022 AMC 12B 第 24 题

A chord between vertices $d$ apart has squared length $2 - 2\cos\dfrac{2\pi d}{7};$ there are $7$ of each type $d = 1, 2, 3$

Use $\sum_{d=1}^{3} \cos\dfrac{2\pi d}{7} = -\dfrac12$ to evaluate $\sum \left(2 - 2\cos\dfrac{2\pi d}{7}\right)^2$

A chord joining two vertices $d$ steps apart has squared length $2 - 2\cos\dfrac{2\pi d}{7},$ and there are $7$ chords for each of $d = 1, 2, 3.$ The required sum is $$7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2.$$

Using $\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12$ and $\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54,$ the inner sum expands to $4\left(3 + 1 + \tfrac54\right) = 21.$

Therefore the total is $7 \cdot 21 = 147.$

Thus, the correct answer is C.