2024 AMC 12A Problems/Problem 23
Problem
What is the value of
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 (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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