2024 AMC 12A Problem 23

Question

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}

28

68

70

72

84

Let

a,b,c,d=\tan^2\frac{\pi}{16},

\tan^2\frac{3\pi}{16},

\tan^2\frac{5\pi}{16},

\tan^2\frac{7\pi}{16};

the sum factors as

(a+d)(b+c)

\tan\frac{7\pi}{16}=\cot\frac{\pi}{16},

and

\tan^2x+\cot^2x=\dfrac{4}{\sin^2 2x}-2

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},

so

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.

Problem 23 in other years