AMC12 2024 A

First, notice that \[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}\] \[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\] Here, we make use of the fact that \[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\] \[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\] \[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{1}{\cos x \sin x}\right)^2-2\] \[=\left(\frac{2}{\sin 2x}\right)^2-2\] \[=\frac{4}{\sin^2 2x}-2\] Hence, \[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\] \[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\] Note that \[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\] \[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\] Hence, \[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\] \[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\] \[=(14+8\sqrt{2})(14-8\sqrt{2})\] \[=68\] Therefore, the answer is $\fbox{\textbf{(B) } 68}$.