2024 AMC 12A Problems/Problem 23

Revision as of 19:56, 8 November 2024 by Tsun26 (talk | contribs) (Solution 1 (Trigonometric Identities))

Problem

What is the value of \[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]

$\textbf{(A) } 28 \qquad \textbf{(B) } 68 \qquad \textbf{(C) } 70 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 84$

Solution 1 (Trigonometric Identities)

First, notice that \[ \tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{3\pi}{16} + \tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{5\pi}{16} + \tan^2 \frac{3\pi}{16} \cdot \tan^2 \frac{7\pi}{16} + \tan^2 \frac{5\pi}{16} \cdot \tan^2 \frac{7\pi}{16} \] can be rewritten as \[ = \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right) \]

Here, we use the identity \[ \tan^2 x + \tan^2 \left( \frac{\pi}{2} - x \right) = \left( \tan x + \tan \left( \frac{\pi}{2} - x \right) \right)^2 - 2 \] to rewrite this as \[ = \left( \frac{\sin x}{\cos x} + \frac{\sin \left( \frac{\pi}{2} - x \right)}{\cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \] \[ = \left( \frac{\sin x \cos \left( \frac{\pi}{2} - x \right) + \sin \left( \frac{\pi}{2} - x \right) \cos x}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \] \[ = \left( \frac{\sin \frac{\pi}{2}}{\cos x \cos \left( \frac{\pi}{2} - x \right)} \right)^2 - 2 \] \[ = \left( \frac{1}{\cos x \sin x} \right)^2 - 2 \] \[ = \left( \frac{2}{\sin 2x} \right)^2 - 2 \] \[ = \frac{4}{\sin^2 2x} - 2 \]

Hence, \[ \left( \tan^2 \frac{\pi}{16} + \tan^2 \frac{7\pi}{16} \right) \cdot \left( \tan^2 \frac{3\pi}{16} + \tan^2 \frac{5\pi}{16} \right) = \left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right) \]

Note that \[ \sin^2 \frac{\pi}{8} = \frac{1 - \cos \frac{\pi}{4}}{2} = \frac{2 - \sqrt{2}}{4} \] and \[ \sin^2 \frac{3\pi}{8} = \frac{1 - \cos \frac{3\pi}{4}}{2} = \frac{2 + \sqrt{2}}{4} \]

Thus, \[ \left( \frac{4}{\sin^2 \frac{\pi}{8}} - 2 \right) \left( \frac{4}{\sin^2 \frac{3\pi}{8}} - 2 \right) = \left( \frac{16}{2 - \sqrt{2}} - 2 \right) \left( \frac{16}{2 + \sqrt{2}} - 2 \right) \]

Simplifying, \[ = (14 + 8\sqrt{2})(14 - 8\sqrt{2}) = 68 \]

Therefore, the answer is \(\boxed{\textbf{(B) } 68}\).

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png