What is the value of \tan^2(π)/(16)×\tan^2(3π)/(16)+\tan^2(π)/(16)×\tan^2(5π)/(16)+\tan^2(3π)/(16)×\tan^2(7π)/(16)+\tan^2(5π)/(16)×\tan^2(7π)/(16)?

2024 AMC 12A Problem 23 — Solution

2024 AMC 12A Problem 23

Below is the professionally curated solution for Problem 23 of the 2024 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

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Concepts:trigonometric identity factoring

Difficulty rating: 2370

23.

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

$28$

$68$

$70$

$72$

$84$

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 $d=\cot^2\tfrac{\pi}{16},$ so $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}=\tfrac{4}{\sin^2(\pi/8)}-2=14+8\sqrt2.$ Likewise $b+c=\tfrac{4}{\sin^2(3\pi/8)}-2=14-8\sqrt2.$ Their product is $14^2-(8\sqrt2)^2=196-128=68.$

Thus, the correct answer is B.

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