2024 AMC 12A Problem 23

Problem 23 of 25HarderAlgebra Geometry

What is the value of

\begin{aligned} &\tan^2\frac{\pi}{16}\cdot\tan^2\frac{3\pi}{16} \\ &\quad {}+\tan^2\frac{\pi}{16}\cdot\tan^2\frac{5\pi}{16} \\ &\quad {}+\tan^2\frac{3\pi}{16}\cdot\tan^2\frac{7\pi}{16} \\ &\quad {}+\tan^2\frac{5\pi}{16}\cdot\tan^2\frac{7\pi}{16}? \end{aligned}

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Solution

With

a=\tan^2\tfrac{\pi}{16},\

b=\tan^2\tfrac{3\pi}{16},\

c=\tan^2\tfrac{5\pi}{16},\

d=\tan^2\tfrac{7\pi}{16},

the expression is

ab+ac+bd+cd

=(a+d)(b+c).

Since

\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},

we have

d=\cot^2\tfrac{\pi}{16},

a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}

=\tfrac{4}{\sin^2(\frac{\pi}{8})}-2

=14+8\sqrt2.

Likewise

b+c=\tfrac{4}{\sin^2(\frac{3\pi}{8})}-2

=14-8\sqrt2.

Their product is

14^2-(8\sqrt2)^2=196-128=68.

Thus, the correct answer is B.