AMC 12 Mock Exam 1 Problem 20

Let $P$ be the requested product and $\omega=e^{2\pi i/11}$. Since $\cos\frac{2k\pi}{11}=\cos\frac{2(11-k)\pi}{11}$, the product of $\left(3-2\cos\frac{2k\pi}{11}\right)$ over $k=1,\dots,10$ equals $P^2$, and $P>0$ because every factor exceeds $1$.

Now $3-2\cos\frac{2k\pi}{11}=3-\omega^k-\omega^{-k}=-\omega^{-k}\left(\omega^{2k}-3\omega^k+1\right)=-\omega^{-k}(\omega^k-\alpha)(\omega^k-\beta)$, where $\alpha,\beta$ are the roots of $t^2-3t+1$, so $\alpha+\beta=3$ and $\alpha\beta=1$. Multiply over $k=1,\dots,10$. First, $\prod(-\omega^{-k})=(-1)^{10}\omega^{-55}=1$. Second, since $\prod_{k=1}^{10}(x-\omega^k)=\frac{x^{11}-1}{x-1}$ and there are ten factors (an even number), $\prod_k(\omega^k-\alpha)=\frac{\alpha^{11}-1}{\alpha-1}$, and similarly for $\beta$. Hence $$P^2=\frac{(\alpha^{11}-1)(\beta^{11}-1)}{(\alpha-1)(\beta-1)}=\frac{(\alpha\beta)^{11}-(\alpha^{11}+\beta^{11})+1}{\alpha\beta-(\alpha+\beta)+1}=\frac{2-(\alpha^{11}+\beta^{11})}{-1}.$$ Let $s_n=\alpha^n+\beta^n$. Then $s_0=2$, $s_1=3$, and $s_n=3s_{n-1}-s_{n-2}$, giving $$7,\ 18,\ 47,\ 123,\ 322,\ 843,\ 2207,\ 5778,\ 15127,\ 39603\qquad(n=2,\dots,11).$$ So $P^2=39603-2=39601=199^2$ and $P=\boxed{199}$.

(Several distractors come from the same calculation: $123$ and $322$ are $s_5$ and $s_6$, and $89$ and $233$ are Fibonacci numbers.)