AMC 12 Mock Exam 1 Problem 24

Let $n=10$ and $\zeta_k=e^{2\pi ik/n}$ for $k=1,\dots,n-1$. For $\theta=\frac{2\pi k}{n}$, $$w_k=\frac{1}{1-\zeta_k}=\frac{1}{-2i\sin\frac\theta2\,e^{i\theta/2}}=\frac12+\frac i2\cot\frac{k\pi}{n}.$$ Since $\zeta_k=1-\frac1{w_k}$ and $\zeta_k^n=1$, each $w_k$ is a root of $(w-1)^n=w^n$, i.e., of $$w^n-(w-1)^n=n\,w^{n-1}-\tbinom n2w^{n-2}+\tbinom n3w^{n-3}-\cdots=0,$$ a polynomial of degree $n-1$ whose roots are exactly $w_1,\dots,w_{n-1}$. By Vieta, $\sum w_k=\frac{\binom n2}{n}=\frac{n-1}{2}$ and $\sum_{j<k}w_jw_k=\frac{\binom n3}{n}=\frac{(n-1)(n-2)}6$, so $$\sum w_k^2=\frac{(n-1)^2}{4}-\frac{(n-1)(n-2)}{3}.$$ Now $\operatorname{Re}(w_k^2)=\frac14\left(1-\cot^2\frac{k\pi}n\right)$, so taking real parts, $$\frac{n-1}{4}-\frac14\sum\cot^2\tfrac{k\pi}n=\frac{(n-1)^2}{4}-\frac{(n-1)(n-2)}{3},$$ which gives $\sum_{k=1}^{n-1}\cot^2\frac{k\pi}{n}=\frac{(n-1)(n-2)}{3}$. Since $\csc^2=1+\cot^2$, $$\sum_{k=1}^{n-1}\csc^2\tfrac{k\pi}{n}=(n-1)+\frac{(n-1)(n-2)}3=\frac{n^2-1}{3}.$$ For $n=10$ this is $\frac{99}{3}=\boxed{33}$.