Difference between revisions of "2024 AMC 12A Problems/Problem 23"

Latest revision as of 10:28, 8 September 2025

1 Problem
2 Solution 1 (Trigonometric Identities)
3 Solution 2 (Another Identity)
4 Solution 3 (Complex Numbers)
5 Solution 5 (Transformation)
6 Solution 6 (Half angle formula twice)
7 Solution 7 (Find each individual tan)
8 Solution 8 (single formula)
9 Solution 9 (Vietas)
10 Alternate proof of the two tangent squares formula
11 Solution 10 (Options)
12 Video Solution
13 See also

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

$\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}$

$=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})$

Here, we make use of the fact that

$\tan^2 x+\tan^2 (\frac{\pi}{2}-x)$ $=(\tan x+\tan (\frac{\pi}{2}-x))^2-2$ $=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2$ $=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2$ $=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2$ $=\left(\frac{1}{\cos x \sin x}\right)^2-2$ $=\left(\frac{2}{\sin 2x}\right)^2-2$ $=\frac{4}{\sin^2 2x}-2$

Hence,

$(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})$ $=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)$

Note that

$\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}$

$\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}$

Hence,

$\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)$

$=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)$

$=(14+8\sqrt{2})(14-8\sqrt{2})$

$=68$

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

~tsun26

Solution 2 (Another Identity)

First, notice that

$\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}$

$=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})$

Here, we make use of the fact that

$\begin{align*} \tan^2 x+\tan^2 (\frac{\pi}{2}-x) &= (\tan x - \tan (\frac{\pi}{2} - x))^2 + 2\\ &= (\tan (\frac{\pi}{2} - 2x) \cdot (1 + \tan x \tan (\frac{\pi}{2} - x))^2 + 2~~~~(\mathrm{difference~of~two~tan})\\ &= (\tan (\frac{\pi}{2} - 2x) \cdot (1 + 1))^2 + 2\\ &= 4\tan^2 (\frac{\pi}{2} - 2x) + 2 \end{align*}$

Hence,

$\begin{align*} (\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}) &= (4\tan^2 (\frac{\pi}{2} - \frac{\pi}{16} \cdot 2) + 2)(4\tan^2 (\frac{\pi}{2} - \frac{3\pi}{16} \cdot 2) + 2)\\ &= (4\tan^2 \frac{3\pi}{8} + 2)(4\tan^2 \frac{\pi}{8} + 2)\\ &= 16\tan^2 \frac{3\pi}{8} \cdot \tan^2 \frac{\pi}{8} + 8(\tan^2 \frac{3\pi}{8} + \tan^2 \frac{\pi}{8}) + 4\\ &= 16 + 8(4\tan^2 (\frac{\pi}{2} - \frac{\pi}{8} \cdot 2) + 2) + 4\\ &= 16 + 8(4\tan^2 \frac{\pi}{4} + 2) + 4\\ &= 16 + 8(4 + 2) + 4\\ &= 68 \end{align*}$

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

~reda_mandymath

Solution 3 (Complex Numbers)

Let $\theta = \frac{\pi}{16}$ . Then, $y = e^{8i\theta} = e^{\frac{\pi}{2} i} = (\cos \theta + i\sin \theta)^8 = 0 + i.$ Expanding by using a binomial expansion, $\Re(y) = \cos^8 \theta - 28 \cos^6 \theta \sin^2 \theta + 70 \cos^4 \theta \sin^4 \theta - 28 \cos^2 \theta \sin^6 \theta + \sin^8\theta =0.$ Divide by $\cos^8 \theta$ and notice we can set $\frac{\sin \theta}{\cos \theta} = x$ where $x = \tan(\theta)$ . Then, define $f(x)$ so that $f(x) = 1 - 28 x^2 + 70 x^4 - 28 x^6 + x^8.$

Notice that we can have $(\cos \theta_k + i \sin \theta_k)^8 = 0 \pm i$ because we are only considering the real parts. We only have this when $k \equiv 1,3 \mod 4$ , meaning $k \equiv 1 \mod 2$ . This means that we have $k = 1,3,5,7,9,11,13,15$ as unique roots (we get them from $k\theta \in [0,\pi]$ ) and by using the fact that $\tan(\pi - \theta) = -\tan \theta$ , we get $x \in \left\{\tan \theta, -\tan \theta, \tan \left(3 \theta \right), -\tan \left(3 \theta \right), \tan \left(5 \theta \right), -\tan \left(5 \theta \right), \tan \left(7 \theta \right), -\tan \left(7 \theta \right) \right\}$ Since we have a monic polynomial, by the Fundamental Theorem of Algebra, $f(x) = (x-\tan \theta)(x+\tan \theta) (x-\tan \left(3 \theta \right))(x+\tan \left(3 \theta \right)) (x-\tan \left(5 \theta \right))(x+\tan \left(5 \theta \right))(x-\tan\left(7 \theta \right))(x+\tan \left(7 \theta \right))$ $f(x) = (x^2 - \tan^2 \theta)(x^2 - \tan^2 (3\theta))(x^2 - \tan^2 (5\theta))(x^2 - \tan^2 (7\theta))$ Looking at the $x^4$ term in the expansion for $f(x)$ and using vietas gives us $\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 \theta \tan^2 (7\theta) + \tan^2 (3\theta) \tan^2 (5\theta)$ $+ \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) = \frac{70}{1} = 70.$ Since $\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta$ and $\tan \theta \cot \theta = 1$ $\tan^2 \theta \tan^2 (7\theta) = \tan^2 (3\theta) \tan^2 (5\theta) = 1.$ Therefore $\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) + 2 = 70.$ $\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) = \boxed{\textbf{(B) } 68}$

~KEVIN_LIU

Solution 5 (Transformation)

Set x = $\pi/16$ , 7x = $\pi/2$ - x , set C7 = $cos^2(7x)$ , C5 = $cos^2(5x)$ , C3 = $cos^2(3x)$ , C= $cos^2(x)$ , S2 = $sin^2(2x)$ , S6 = $sin^2(6x), etc.$

First, notice that $\tan^2 x \cdot \tan^2 3x + \tan^2 3x \cdot \tan^2 5x+\tan^2 3x \cdot \tan^2 7x+\tan^2 5x \cdot \tan^2 7x$ $=(\tan^2x+\tan^2 7x)(\tan^23x+\tan^2 5x)$ $=(\frac{1}{C} - 1 +\frac{1}{C7}-1)(\frac{1}{C3} - 1 +\frac{1}{C5}-1)$ $=(\frac{C+C7}{C \cdot C7} -2)( \frac{C3+C5}{C3 \cdot C5} -2)$ $=(\frac{1}{C \cdot S} -2)( \frac{1}{C3 \cdot S3} -2)$ $=(\frac{4}{S2} -2)( \frac{4}{S6} -2)$ $=4(\frac{2-S2}{S2})( \frac{2-S6}{S6})$ $=4(\frac{4-2 \cdot S2-S \cdot S6 }{S2 \cdot S6}+1)$ $=4 + \frac{8}{S2 \cdot S6}$ $=4 + \frac{32}{S4}$ $=4 + 64$ $= 68$

~luckuso

Solution 6 (Half angle formula twice)

So from the question we have: $\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}$

$=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})$

Using $\tan^2\frac{\theta}{2}=\frac{1-\cos\theta}{1+\cos\theta}$

$=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1+\cos\frac{7\pi}{8}}{1-\cos\frac{7\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1+\cos\frac{5\pi}{8}}{1-\cos\frac{5\pi}{8}})$

Using $\cos\theta=-\cos(\pi-\theta)$

$=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1-\cos\frac{\pi}{8}}{1+\cos\frac{\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1-\cos\frac{3\pi}{8}}{1+\cos\frac{3\pi}{8}})$

$=(\frac{(1+\cos\frac{\pi}{8})^2+(1-\cos\frac{\pi}{8})^2}{1-\cos^2\frac{\pi}{8}})(\frac{(1+\cos\frac{3\pi}{8})^2+(1-\cos\frac{3\pi}{8})^2}{1-\cos^2\frac{3\pi}{8}})$

$=(\frac{2+2\cos^2\frac{\pi}{8}}{1-\cos^2\frac{\pi}{8}})(\frac{2+2\cos^2\frac{3\pi}{8}}{1-\cos^2\frac{3\pi}{8}})$

Using $\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}$

$=(\frac{2+1+\cos\frac{\pi}{4}}{1-\frac{1+\cos\frac{\pi}{4}}{2}})(\frac{2+1+\cos\frac{3\pi}{4}}{1-\frac{1+\cos\frac{3\pi}{4}}{2}})$

$=(\frac{12+2\sqrt{2}}{4-2\sqrt{2}})(\frac{12-2\sqrt{2}}{4+2\sqrt{2}})$

$=\frac{136}{2}=\boxed{\textbf{B) }68 }$

~ERiccc

Solution 7 (Find each individual tan)

The half angle formula for $\tan^2$ is $\tan^2\frac{\theta}{2} = \frac{1 - \cos\theta}{1 + \cos\theta}$ and the half angle formula for cosine is $\cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}}.$ We can use this to find each tan:

$\cos(\pi/8) = \sqrt{\frac{1 + \cos(\pi/4)}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\tan^2(\pi/16) = \frac{1 - \cos(\pi/8)}{1 + \cos(\pi/8)} = \frac{2 - \sqrt{2 + \sqrt{2}}}{2 + \sqrt{2 + \sqrt{2}}}$

$\cos(3\pi/8) = \sqrt{\frac{1 + \cos(3\pi/4)}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$ $\tan^2(3\pi/16) = \frac{1 - \cos(3\pi/8)}{1 + \cos(3\pi/8)} = \frac{2 - \sqrt{2 - \sqrt{2}}}{2 + \sqrt{2 - \sqrt{2}}}$

$\cos(5\pi/8) = - \sqrt{\frac{1 + \cos(5\pi/4)}{2}} = -\frac{\sqrt{2 - \sqrt{2}}}{2}$ $\tan^2(5\pi/16) = \frac{1 - \cos(5\pi/8)}{1 + \cos(5\pi/8)} = \frac{2 + \sqrt{2 - \sqrt{2}}}{2 - \sqrt{2 - \sqrt{2}}}$

$\cos(7\pi/8) = - \sqrt{\frac{1 + \cos(7\pi/4)}{2}} = -\frac{\sqrt{2 + \sqrt{2}}}{2}$ $\tan^2(7\pi/16) = \frac{1 - \cos(7\pi/8)}{1 + \cos(7\pi/8)} = \frac{2 + \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2 + \sqrt{2}}}$

The problem's expression can be factored as $(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}).$ So the answer is

$(\frac{2 - \sqrt{2 + \sqrt{2}}}{2 + \sqrt{2 + \sqrt{2}}} + \frac{2 + \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2 + \sqrt{2}}}) \cdot (\frac{2 - \sqrt{2 - \sqrt{2}}}{2 + \sqrt{2 - \sqrt{2}}} + \frac{2 + \sqrt{2 - \sqrt{2}}}{2 - \sqrt{2 - \sqrt{2}}}) =$

$(\frac{12 + 2\sqrt{2}}{2 - \sqrt{2}}) \cdot (\frac{12 - 2\sqrt{2}}{2 + \sqrt{2}}) =$

$(14 + 8\sqrt{2}) \cdot (14 - 8\sqrt{2}) =$

$196 - 128 = \boxed{68}.$

~grogg007

Solution 8 (single formula)

$\cot \alpha - \tan \alpha = 2 \cot 2 \alpha \implies \cot^2 \alpha + \tan^2 \alpha = 4 \cot^2 2 \alpha + 2.$ We use $\alpha = \frac {\pi}{16}$ for $(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}).$

$(\tan^2 \alpha + \cot^2 \alpha)(\tan^2 (\frac{\pi}{4} - \alpha) + \cot^2 (\frac{\pi}{4} - \alpha)) = (4 \cot^2 2 \alpha + 2)(4 \cot^2 (\frac{\pi}{2} - 2\alpha) +2) =$ $= 4 \cdot(4+ 2\tan^2 2\alpha + 2\cot^2 2\alpha +1) = 20 + 8 \cdot (4 \cot^2 4 \alpha +2) = 68.\blacksquare$ vladimir.shelomovskii@gmail.com, vvsss

Solution 9 (Vietas)

As the above solutions noted, we can factor the expression into $(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})$ .

Before we directly solve this problem, let's analyze the roots of $\tan(4\tan^{-1}{x}) = 1$ , or equivalently using tangent expansion formula, $\frac{1-6x^2+x^4}{4x-4x^3}=1$ , which implies $x^4+4x^3-6x^2-4x+1=0$ . Now note that the roots of this equation are precisely $\tan\frac{\pi}{16}, \tan\frac{5\pi}{16}, \tan\frac{9\pi}{16}, \tan\frac{13\pi}{16}$ , so the second symmetric sum of these four numbers is $6$ by Vieta's. Thus, we have $\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6$ Upon further inspection, $\tan\frac{\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{13\pi}{16}=-2$ using the fact that $\tan(x)*\tan(x + \pi/2) = -1$ . Hence, we have $\tan\frac{\pi}{16}\tan\frac{5\pi}{16}-1+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}-1+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6$ $\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=8$ $(\tan\frac{\pi}{16}+\tan\frac{9\pi}{16})(\tan\frac{5\pi}{16}+\tan\frac{13\pi}{16})=8$

Now, we return to the problem statement, where we see a similar squared sum. We use this motivation to square our equation above to obtain

$(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16}-2)(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16}-2)=64$ $(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})+4=64$ $(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60$ Then, use the fact that $\tan^2{x}=\tan^2{\pi/2-x}$ to get $(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60$ Hold on; the first term is exactly what we are solving for! It thus suffices to find $\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16}$ . Fortunately, this is just ${S_1}^2-2{S_2}$ (Where $S_n$ is the nth symmetric sum), with relation to roots of $x^4+4x^3-6x^2-4x+1=0$ . By Vieta's, this is just $(-4)^2-2(6)=4$ .

Finally, we plug this value into our equation to obtain $(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(4)=60$ $(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}$

Alternate proof of the two tangent squares formula

We want to simplify $\tan^{2}(x)$ + $\tan^{2}(\frac{\pi}{2} - x)$ . We make use of the fact that $\tan(\frac{\pi}{2} - x)$ = $\cot(x)$ .Then, the expression becomes $\tan^{2}(x)$ + $\cot^{2}(x)$ . Notice we can write: $(\tan x + \cot x)^{2}$ as $\tan^{2}(x)$ + $\cot^{2}(x)$ + 2 as tangent and cotangent are reciprocals of each other. Then, the sum of the tangent and cotangent can be simplified to $\frac{\sec^{2}{x}}{\tan x}$ . Using the fact that secant is the reciprocal of cosine and tangent is the ratio of sine and cosine, we can simplify that expression to $\frac{1}{\sin x \cos x}$ . So, we have that: $\tan^{2}(x)$ + $\cot^{2}(x)$ = $\frac{1}{(\sin x \cos x)^2} - {2}$ which can be simplfied to: $2\Bigr(\frac{2}{\sin^{2}(2x)} - 1\Bigr)$ or $\frac{4}{\sin^{2}(2x)} - 2$ as stated in earlier solutions.

~ilikemath247365

Solution 10 (Options)

For this question, there are five options: $28$, $68$, $70$, $72$, and $84$. Since $28$ is too small and $84$ is too large, these two options can be eliminated. At this point, only three options remain. While one could make a guess here, it is better to analyze these remaining options further. When we divide each of these three options by $2$, we get $34$, $35$, and $36$ respectively. Option C ($70$) should be eliminated because, after division by $2$, it is the only odd number among the three results. Last but not least, Option D ($72$) appears excessively frequently in AMC answer choices. Therefore, it is reasonable to conjecture that the Mathematical Association of America (MAA) will avoid using this option again for AMC, leading to the elimination of D. As a result, the only remaining option is B ($68$).

Video Solution

2024 AMC 12 A #23

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2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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Difference between revisions of "2024 AMC 12A Problems/Problem 23"

Latest revision as of 10:28, 8 September 2025

Contents

Problem

Solution 1 (Trigonometric Identities)

Solution 2 (Another Identity)

Solution 3 (Complex Numbers)

Solution 5 (Transformation)

Solution 6 (Half angle formula twice)

Solution 7 (Find each individual tan)

Solution 8 (single formula)

Solution 9 (Vietas)

Alternate proof of the two tangent squares formula

Solution 10 (Options)

Video Solution

See also

@@ Line 1: / Line 1: @@
-{{duplicate|[[2024 AMC 12A Problems/Problem 23|2024 AMC 12A #23]] and [[2024 AMC 10A Problems/Problem 25|2024 AMC 10A #25]]}}
+==Problem==
+What is the value of <cmath>\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}?</cmath>
+<math>\textbf{(A) } 28 \qquad \textbf{(B) } 68 \qquad \textbf{(C) } 70 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 84</math>
+==Solution 1 (Trigonometric Identities)==
+First, notice that
+<cmath>\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}</cmath>
+<cmath>=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})</cmath>
+Here, we make use of the fact that
+<cmath>\tan^2 x+\tan^2 (\frac{\pi}{2}-x)</cmath>
+<cmath>=(\tan x+\tan (\frac{\pi}{2}-x))^2-2</cmath>
+<cmath>=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
+<cmath>=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
+<cmath>=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2</cmath>
+<cmath>=\left(\frac{1}{\cos x \sin x}\right)^2-2</cmath>
+<cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath>
+<cmath>=\frac{4}{\sin^2 2x}-2</cmath>
+Hence,
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})</cmath>
+<cmath>=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath>
+Note that
+<cmath>\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}</cmath>
+<cmath>\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}</cmath>
+Hence,
+<cmath>\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath>
+<cmath>=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)</cmath>
+<cmath>=(14+8\sqrt{2})(14-8\sqrt{2})</cmath>
+<cmath>=68</cmath>
+Therefore, the answer is <math>\fbox{\textbf{(B) } 68}</math>.
+~tsun26
+==Solution 2 (Another Identity)==
+First, notice that
+<cmath>\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}</cmath>
-==Problem==
-In parallelogram <math>ABCD</math>, let <math>\omega</math> be the circle with diameter <math>\overline{AD}</math> and suppose <math>P</math> and <math>Q</math> are points on <math>\omega</math> such that both lines <math>BP</math> and <math>BQ</math> are tangent to <math>\omega</math>. If <math>BC = 8</math>, <math>BP = 3</math>, and line <math>PQ</math> bisects <math>\overline{CD}</math>, what is <math>AC^{2}</math>?
-<math>\textbf{(A)}~180\qquad\textbf{(B)}~181\qquad\textbf{(C)}~182\qquad\textbf{(D)}~183\qquad\textbf{(E)}~184</math>
+<cmath>=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})</cmath>
+Here, we make use of the fact that
+<cmath>
+\begin{align*}
+\tan^2 x+\tan^2 (\frac{\pi}{2}-x) &= (\tan x - \tan (\frac{\pi}{2} - x))^2 + 2\\
+&= (\tan (\frac{\pi}{2} - 2x) \cdot (1 + \tan x \tan (\frac{\pi}{2} - x))^2 + 2~~~~(\mathrm{difference~of~two~tan})\\
+&= (\tan (\frac{\pi}{2} - 2x) \cdot (1 + 1))^2 + 2\\
+&= 4\tan^2 (\frac{\pi}{2} - 2x) + 2
+\end{align*}
+</cmath>
+Hence,
+<cmath>
+\begin{align*}
+(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}) &= (4\tan^2 (\frac{\pi}{2} - \frac{\pi}{16} \cdot 2) + 2)(4\tan^2 (\frac{\pi}{2} - \frac{3\pi}{16} \cdot 2) + 2)\\
+&= (4\tan^2 \frac{3\pi}{8} + 2)(4\tan^2 \frac{\pi}{8} + 2)\\
+&= 16\tan^2 \frac{3\pi}{8} \cdot \tan^2 \frac{\pi}{8} + 8(\tan^2 \frac{3\pi}{8} + \tan^2 \frac{\pi}{8}) + 4\\
+&= 16 + 8(4\tan^2 (\frac{\pi}{2} - \frac{\pi}{8} \cdot 2) + 2) + 4\\
+&= 16 + 8(4\tan^2 \frac{\pi}{4} + 2) + 4\\
+&= 16 + 8(4 + 2) + 4\\
+&= 68
+\end{align*}
+</cmath>
+Therefore, the answer is <math>\fbox{\textbf{(B) } 68}</math>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
+==Solution 3 (Complex Numbers)==
+Let <math>\theta = \frac{\pi}{16}</math>. Then,
+<cmath>
+y = e^{8i\theta} = e^{\frac{\pi}{2} i} = (\cos \theta + i\sin \theta)^8 = 0 + i.
+</cmath>
+Expanding by using a binomial expansion,
+<cmath>
+\Re(y) = \cos^8 \theta - 28 \cos^6 \theta \sin^2 \theta + 70 \cos^4 \theta \sin^4 \theta - 28 \cos^2 \theta \sin^6 \theta +  \sin^8\theta =0.
+</cmath>
+Divide by <math>\cos^8 \theta</math> and notice we can set <math>\frac{\sin \theta}{\cos \theta} = x</math> where <math>x = \tan(\theta)</math>. Then, define <math>f(x)</math> so that
+<cmath>
+f(x) = 1 - 28 x^2 + 70 x^4 - 28 x^6 + x^8.
+</cmath>
+Notice that we can have <math>(\cos \theta_k + i \sin \theta_k)^8 = 0 \pm i</math> because we are only considering the real parts. We only have this when <math>k \equiv 1,3 \mod 4</math>, meaning <math>k \equiv 1 \mod 2</math>. This means that we have <math>k = 1,3,5,7,9,11,13,15</math> as unique roots (we get them from <math>k\theta \in [0,\pi]</math>) and by using the fact that <math>\tan(\pi - \theta) = -\tan \theta</math>, we get <cmath>x \in \left\{\tan \theta, -\tan \theta, \tan \left(3 \theta \right), -\tan \left(3 \theta \right), \tan \left(5 \theta \right), -\tan \left(5 \theta \right), \tan \left(7 \theta \right), -\tan \left(7 \theta \right) \right\} </cmath>
+Since we have a monic polynomial, by the Fundamental Theorem of Algebra,
+<cmath>f(x) = (x-\tan \theta)(x+\tan \theta) (x-\tan \left(3 \theta \right))(x+\tan \left(3 \theta \right)) (x-\tan \left(5 \theta \right))(x+\tan \left(5 \theta \right))(x-\tan\left(7 \theta \right))(x+\tan \left(7 \theta \right))</cmath>
+<cmath>f(x) =  (x^2 - \tan^2 \theta)(x^2 - \tan^2 (3\theta))(x^2 - \tan^2 (5\theta))(x^2 - \tan^2 (7\theta))
+</cmath>
+Looking at the <math>x^4</math> term in the expansion for <math>f(x)</math> and using vietas gives us
+<cmath>
+\tan^2 \theta  \tan^2 (3\theta) + \tan^2 \theta  \tan^2 (5\theta) + \tan^2 \theta  \tan^2 (7\theta) + \tan^2 (3\theta)  \tan^2 (5\theta)
+</cmath>
+<cmath>
++ \tan^2 (3\theta)  \tan^2 (7\theta) + \tan^2 (5\theta)  \tan^2 (7\theta) = \frac{70}{1} = 70.
+</cmath>
+Since <math>\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta</math> and <math> \tan \theta  \cot \theta = 1</math>
+<cmath>
+\tan^2 \theta  \tan^2 (7\theta) = \tan^2 (3\theta)  \tan^2 (5\theta) = 1.
+</cmath>
+Therefore
+<cmath>
+\tan^2 \theta  \tan^2 (3\theta) + \tan^2 \theta  \tan^2 (5\theta) + \tan^2 (3\theta)  \tan^2 (7\theta) + \tan^2 (5\theta)  \tan^2 (7\theta) + 2 = 70.
+</cmath>
+<cmath>
+\tan^2 \theta  \tan^2 (3\theta) + \tan^2 \theta  \tan^2 (5\theta) + \tan^2 (3\theta)  \tan^2 (7\theta) + \tan^2 (5\theta)  \tan^2 (7\theta) = \boxed{\textbf{(B) } 68}
+</cmath>
+~[https://artofproblemsolving.com/wiki/index.php/User:KEVIN_LIU KEVIN_LIU]
+==Solution 5 (Transformation)==
+Set x = <math>\pi/16</math>  , 7x = <math>\pi/2</math> - x ,
+set C7 = <math>cos^2(7x)</math> , C5 = <math>cos^2(5x)</math>, C3 = <math>cos^2(3x)</math>, C= <math>cos^2(x)</math> , S2 = <math>sin^2(2x)</math> , S6 = <math>sin^2(6x), etc.</math>
+First, notice that
+<cmath>\tan^2 x \cdot \tan^2 3x + \tan^2 3x \cdot \tan^2 5x+\tan^2 3x \cdot \tan^2 7x+\tan^2 5x \cdot \tan^2 7x</cmath>
+<cmath>=(\tan^2x+\tan^2 7x)(\tan^23x+\tan^2 5x)</cmath>
+<cmath>=(\frac{1}{C} - 1 +\frac{1}{C7}-1)(\frac{1}{C3} - 1 +\frac{1}{C5}-1)</cmath>
+<cmath>=(\frac{C+C7}{C \cdot C7} -2)( \frac{C3+C5}{C3 \cdot C5} -2)</cmath>
+<cmath>=(\frac{1}{C \cdot S} -2)( \frac{1}{C3 \cdot S3} -2)</cmath>
+<cmath>=(\frac{4}{S2} -2)( \frac{4}{S6} -2)</cmath>
+<cmath>=4(\frac{2-S2}{S2})( \frac{2-S6}{S6})</cmath>
+<cmath>=4(\frac{4-2 \cdot S2-S \cdot S6 }{S2 \cdot S6}+1)</cmath>
+<cmath>=4 + \frac{8}{S2 \cdot S6} </cmath>
+<cmath>=4 + \frac{32}{S4} </cmath>
+<cmath>=4 +  64 </cmath>
+<cmath>= 68 </cmath>
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
+==Solution 6 (Half angle formula twice)==
+So from the question we have:
+<cmath>\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}</cmath>
+<cmath>=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})</cmath>
+Using <math>\tan^2\frac{\theta}{2}=\frac{1-\cos\theta}{1+\cos\theta}</math>
+<cmath>=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1+\cos\frac{7\pi}{8}}{1-\cos\frac{7\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1+\cos\frac{5\pi}{8}}{1-\cos\frac{5\pi}{8}})</cmath>
+Using <math>\cos\theta=-\cos(\pi-\theta)</math>
+<cmath>=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1-\cos\frac{\pi}{8}}{1+\cos\frac{\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1-\cos\frac{3\pi}{8}}{1+\cos\frac{3\pi}{8}})</cmath>
+<cmath>=(\frac{(1+\cos\frac{\pi}{8})^2+(1-\cos\frac{\pi}{8})^2}{1-\cos^2\frac{\pi}{8}})(\frac{(1+\cos\frac{3\pi}{8})^2+(1-\cos\frac{3\pi}{8})^2}{1-\cos^2\frac{3\pi}{8}})</cmath>
+<cmath>=(\frac{2+2\cos^2\frac{\pi}{8}}{1-\cos^2\frac{\pi}{8}})(\frac{2+2\cos^2\frac{3\pi}{8}}{1-\cos^2\frac{3\pi}{8}})</cmath>
+Using <math>\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}</math>
+<cmath>=(\frac{2+1+\cos\frac{\pi}{4}}{1-\frac{1+\cos\frac{\pi}{4}}{2}})(\frac{2+1+\cos\frac{3\pi}{4}}{1-\frac{1+\cos\frac{3\pi}{4}}{2}})</cmath>
+<cmath>=(\frac{12+2\sqrt{2}}{4-2\sqrt{2}})(\frac{12-2\sqrt{2}}{4+2\sqrt{2}})</cmath>
+<cmath>=\frac{136}{2}=\boxed{\textbf{B) }68 }</cmath>
+~ERiccc
+==Solution 7 (Find each individual tan)==
+The half angle formula for <math>\tan^2</math> is <math>\tan^2\frac{\theta}{2} = \frac{1 - \cos\theta}{1 + \cos\theta}</math> and the half angle formula for cosine is <math>\cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}}.</math> We can use this to find each tan:
+<cmath>\cos(\pi/8) = \sqrt{\frac{1 + \cos(\pi/4)}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}</cmath>
+<cmath>\tan^2(\pi/16) = \frac{1 - \cos(\pi/8)}{1 + \cos(\pi/8)} = \frac{2 - \sqrt{2 + \sqrt{2}}}{2 + \sqrt{2 + \sqrt{2}}}</cmath>
+<cmath>\cos(3\pi/8) = \sqrt{\frac{1 + \cos(3\pi/4)}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}</cmath>
+<cmath>\tan^2(3\pi/16) = \frac{1 - \cos(3\pi/8)}{1 + \cos(3\pi/8)} = \frac{2 - \sqrt{2 - \sqrt{2}}}{2 + \sqrt{2 - \sqrt{2}}}</cmath>
+<cmath>\cos(5\pi/8) = - \sqrt{\frac{1 + \cos(5\pi/4)}{2}} = -\frac{\sqrt{2 - \sqrt{2}}}{2}</cmath>
+<cmath>\tan^2(5\pi/16) = \frac{1 - \cos(5\pi/8)}{1 + \cos(5\pi/8)} = \frac{2 + \sqrt{2 - \sqrt{2}}}{2 - \sqrt{2 - \sqrt{2}}}</cmath>
+<cmath>\cos(7\pi/8) = - \sqrt{\frac{1 + \cos(7\pi/4)}{2}} = -\frac{\sqrt{2 + \sqrt{2}}}{2}</cmath>
+<cmath>\tan^2(7\pi/16) = \frac{1 - \cos(7\pi/8)}{1 + \cos(7\pi/8)} = \frac{2 + \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2 + \sqrt{2}}}</cmath>
+The problem's expression can be factored as <cmath>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}).</cmath> So the answer is
+<cmath>(\frac{2 - \sqrt{2 + \sqrt{2}}}{2 + \sqrt{2 + \sqrt{2}}} + \frac{2 + \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2 + \sqrt{2}}}) \cdot (\frac{2 - \sqrt{2 - \sqrt{2}}}{2 + \sqrt{2 - \sqrt{2}}} + \frac{2 + \sqrt{2 - \sqrt{2}}}{2 - \sqrt{2 - \sqrt{2}}}) = </cmath>
+<cmath>(\frac{12 + 2\sqrt{2}}{2 - \sqrt{2}}) \cdot (\frac{12 - 2\sqrt{2}}{2 + \sqrt{2}}) = </cmath>
+<cmath>(14 + 8\sqrt{2}) \cdot (14 - 8\sqrt{2}) = </cmath>
+<cmath>196 - 128 = \boxed{68}.</cmath>
+~[[User:grogg007|grogg007]]
+==Solution 8 (single formula)==
+<cmath>\cot \alpha - \tan \alpha = 2 \cot 2 \alpha \implies \cot^2 \alpha + \tan^2 \alpha = 4 \cot^2 2 \alpha + 2.</cmath>
+We use <math>\alpha = \frac {\pi}{16}</math> for <math>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16}).</math>
+<cmath>(\tan^2 \alpha + \cot^2 \alpha)(\tan^2 (\frac{\pi}{4} - \alpha) + \cot^2 (\frac{\pi}{4} - \alpha)) = (4 \cot^2 2 \alpha + 2)(4 \cot^2 (\frac{\pi}{2} - 2\alpha) +2) =</cmath>
+<cmath>= 4 \cdot(4+ 2\tan^2 2\alpha + 2\cot^2 2\alpha +1) = 20 + 8 \cdot (4 \cot^2 4 \alpha +2) = 68.\blacksquare</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 9 (Vietas)==
+As the above solutions noted, we can factor the expression into <math>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})</math>.
+Before we directly solve this problem, let's analyze the roots of <math>\tan(4\tan^{-1}{x}) = 1</math>, or equivalently using tangent expansion formula, <math>\frac{1-6x^2+x^4}{4x-4x^3}=1</math>, which implies <math>x^4+4x^3-6x^2-4x+1=0</math>. Now note that the roots of this equation are precisely <math>\tan\frac{\pi}{16}, \tan\frac{5\pi}{16}, \tan\frac{9\pi}{16}, \tan\frac{13\pi}{16}</math>, so the second symmetric sum of these four numbers is <math>6</math> by Vieta's. Thus, we have <cmath>\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6</cmath>
+Upon further inspection, <math>\tan\frac{\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{13\pi}{16}=-2</math> using the fact that <math>\tan(x)*\tan(x + \pi/2) = -1</math>. Hence, we have <cmath>\tan\frac{\pi}{16}\tan\frac{5\pi}{16}-1+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}-1+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6</cmath>
+<cmath>\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=8</cmath>
+<cmath>(\tan\frac{\pi}{16}+\tan\frac{9\pi}{16})(\tan\frac{5\pi}{16}+\tan\frac{13\pi}{16})=8</cmath>
+Now, we return to the problem statement, where we see a similar squared sum. We use this motivation to square our equation above to obtain
-==Solution 1==
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16}-2)(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16}-2)=64</cmath>
-Let <math>M</math>, <math>O</math>, and <math>N</math> be the midpoints of <math>\overline{CD}</math>, <math>\overline{AD}</math>, and <math>\overline{AB}</math>, respectively. Let <math>\overline{PQ}</math> intersect <math>\overline{BO}</math> at <math>R</math> and note that the radius of <math>\omega</math> is <math>\tfrac{AD}{2} = \tfrac{8}{2} = 4</math>. The Pythagorean theorem applied to <math>\triangle BPO</math> gives <math>BO = 5</math>, and the similarity <math>\triangle BPR \sim \triangle BOP</math> implies <math>BR = \tfrac{BP^{2}}{BO} = \tfrac{9}{5}</math>.
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})+4=64</cmath>
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60</cmath>
+Then, use the fact that <math>\tan^2{x}=\tan^2{\pi/2-x}</math> to get
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60</cmath>
+Hold on; the first term is exactly what we are solving for! It thus suffices to find <math>\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16}</math>. Fortunately, this is just <math>{S_1}^2-2{S_2}</math> (Where <math>S_n</math> is the nth symmetric sum), with relation to roots of <math>x^4+4x^3-6x^2-4x+1=0</math>. By Vieta's, this is just <math>(-4)^2-2(6)=4</math>.
-Let <math>\overline{MN}</math> intersect <math>\overline{BO}</math> at point <math>E</math>. Since <math>ABCD</math> is a parallelogram, <math>E</math> is the midpoint of <math>\overline{BO}</math> and <math>EN = \tfrac{AO}{2} = \tfrac{4}{2} = 2</math>. Because <math>MN = BC = 8</math>, we have <math>ME = 6</math> and <math>RE = BE - BR = \tfrac{5}{2} - \tfrac{9}{5} = \tfrac{7}{10}</math>. It follows that <cmath>\cos(\angle MEO) = \cos(\pi - \angle MER) = -\cos(\angle MER) = -\tfrac{7}{60}.</cmath>
+Finally, we plug this value into our equation to obtain
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(4)=60</cmath>
+<cmath>(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}</cmath>
-<asy>
+==Alternate proof of the two tangent squares formula==
-size(7cm);
-defaultpen(fontsize(10pt)+linewidth(0.4));
-pair A = (0, 0), B = (55/12, sqrt(3551)/12), C = (151/12, sqrt(3551)/12), D = (8, 0), M = (C + D)/2, N = (A + B)/2, O = (A + D)/2, E = (B + O)/2, P = intersectionpoints(circle(O,abs(A-O)),circle(E,abs(B-O)/2))[1], Q = intersectionpoints(circle(O,abs(A-O)),circle(E,abs(B-O)/2))[0], R = (P + Q)/2;
+We want to simplify <math>\tan^{2}(x)</math> + <math>\tan^{2}(\frac{\pi}{2} - x)</math>. We make use of the fact that <math>\tan(\frac{\pi}{2} - x)</math> = <math>\cot(x)</math>.Then, the expression becomes <math>\tan^{2}(x)</math> + <math>\cot^{2}(x)</math>.
-draw(A--B--C--D--cycle);
+Notice we can write:
-draw(arc(O,abs(A-O),0,180));
+<math>(\tan x + \cot x)^{2}</math> as <math>\tan^{2}(x)</math> + <math>\cot^{2}(x)</math> + 2 as tangent and cotangent are reciprocals of each other. Then, the sum of the tangent and cotangent can be simplified to <math>\frac{\sec^{2}{x}}{\tan x}</math>. Using the fact that secant is the reciprocal of cosine and tangent is the ratio of sine and cosine, we can simplify that expression to <math>\frac{1}{\sin x \cos x}</math>. So, we have that: <math>\tan^{2}(x)</math> + <math>\cot^{2}(x)</math> = <math>\frac{1}{(\sin x \cos x)^2} - {2}</math> which can be simplfied to: <math>2\Bigr(\frac{2}{\sin^{2}(2x)} - 1\Bigr)</math> or <math>\frac{4}{\sin^{2}(2x)} - 2</math> as stated in earlier solutions.
-draw(B--P, gray);
-draw(B--Q, gray);
-draw(B--O, dashed);
-draw(P--M);
-draw(rightanglemark(M, R, E));
-fill(M--E--O--cycle, pink);
-dot("$A$", A, dir(263));
+~ilikemath247365
-dot("$B$", B, dir(83));
-dot("$C$", C, dir(83));
-dot("$D$", D, dir(263));
-dot("$E$", E, dir(353));
-dot("$M$", M, dir(353));
-dot("$O$", O, dir(263));
-dot("$P$", P, dir(135));
-dot("$Q$", Q, dir(45));
-dot("$R$", R, dir(45));
-dot("$N$", N, dir(187));
-</asy>
-Applying the Law of Cosines in <math>\triangle MEO</math>, <cmath>MO^{2} = 6^{2} + \left(\frac{5}{2}\right)^{2} + 2 \cdot \frac{5}{2} \cdot 6 \cdot \frac{7}{60} = 36 + \frac{25}{4} + \frac{7}{2} = \frac{183}{4}.</cmath> A double homothety at <math>D</math> gives <math>AC^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}</math>.
-==Solution 2==
+==Solution 10 (Options)==
-Let <math>M</math>, <math>O</math>, and <math>N</math> be the midpoints of <math>\overline{CD}</math>, <math>\overline{AD}</math>, and <math>\overline{AB}</math> respectively. Let <math>E</math> be the midpoint of <math>\overline{BO}</math> and let <math>\gamma</math> be the circumcircle of <math>\triangle{BPQ}</math>.
+For this question, there are five options: \(28\), \(68\), \(70\), \(72\), and \(84\). Since \(28\) is too small and \(84\) is too large, these two options can be eliminated. At this point, only three options remain. While one could make a guess here, it is better to analyze these remaining options further. When we divide each of these three options by \(2\), we get \(34\), \(35\), and \(36\) respectively. Option C (\(70\)) should be eliminated because, after division by \(2\), it is the only odd number among the three results. Last but not least, Option D (\(72\)) appears excessively frequently in AMC answer choices. Therefore, it is reasonable to conjecture that the Mathematical Association of America (MAA) will avoid using this option again for AMC, leading to the elimination of D. As a result, the only remaining option is B (\(68\)).
-First, <math>OP = \tfrac{AD}{2} = \tfrac{BC}{2} = 4</math>, so the radius of <math>\omega</math> is <math>4</math>. Since <math>\angle{OPB} = \angle{OQB} = 90^{\circ}</math>, <math>\overline{BO}</math> is a diameter of <math>\gamma</math>, and <math>\gamma</math> has center <math>E</math>. The Pythagorean theorem applied to either <math>\triangle{BPO}</math> or <math>\triangle{BQO}</math> gives that <math>BO = 5</math>, so the radius of <math>\gamma</math> is <math>\tfrac{5}{2}</math>. Since <math>E</math> is the midpoint of <math>\overline{BO}</math>, and <math>ABCD</math> is a parallelogram, we must have that <math>E</math> lies on <math>\overline{MN}</math>, and lies <math>\tfrac{3}{4}</math> of the way from <math>M</math> to <math>N</math>, giving <math>ME = \tfrac{3MN}{4} = 6</math>.
-Since <math>M</math> lies on line <math>PQ</math>, the radical axis of <math>\gamma</math> and <math>\omega</math>, it has equal power with respect to both circles. Thus <cmath>\text{Pow}_{\gamma}{(M)} = ME^{2} - EB^{2} = 6^{2} - \left(\frac{5}{2}\right)^{2} = \frac{119}{4} = \text{Pow}_{\omega}{(M)} = MO^{2} - OA^{2} = MO^{2} - 16,</cmath> and <math>MO^{2} = \tfrac{183}{4}</math>. A double homothety at <math>D</math> finishes, and <math>AC^{2} = (2MO)^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}</math>.
+== Video Solution ==
+[https://youtu.be/Kc8cgdXoOPs?si=rbPKKpKMHMUV6Ta2 2024 AMC 12 A #23]
-==Solution 3==
+[https://mathproblemsolvingskills.wordpress.com/ MathProblemSolvingSkills.com]
-Let <math>M</math> and <math>O</math> be the midpoints of <math>\overline{CD}</math> and <math>\overline{AD}</math>, respectively. Consider taking segment <math>\overline{BO}</math>, translating it <math>8</math> units in the direction of vector <math>\vec{AD}</math>. Then perform a homothety with ratio <math>\tfrac{1}{2}</math> at <math>D</math>, mapping <math>B</math> to <math>M</math> and <math>O</math> to <math>O^{\prime}</math>. Since <math>CO^{\prime} = 2</math> and <math>CO = 4</math>, we have that <math>OO^{\prime} = 6</math>. Also, <math>MO^{\prime} = \tfrac{OB}{2} = \tfrac{5}{2}</math>.
-Let <math>\overline{BO}</math> and <math>\overline{PQ}</math> intersect at <math>R</math>, and note that <math>OR = \tfrac{16}{5}</math>. Since <math>\overline{OR} \perp \overline{PQ}</math>, we also have <math>\overline{O^{\prime}M} \perp \overline{PQ}</math>. Applying the Pythagorean theorem on right trapezoid <math>ORMO^{\prime}</math>, we have <math>RM^{2} = 6^{2} - \left(\tfrac{16}{5} - \tfrac{5}{2}\right)^{2} = \tfrac{3551}{100}</math>. The Pythagorean theorem in <math>\triangle ORM</math> can again be used to calculate <math>MO^{2} = \left(\tfrac{16}{5}\right)^{2} + \tfrac{3551}{100} = \tfrac{183}{4}</math>. As in other solutions, <math>AC^{2} = 4MO^{2} = \boxed{\textbf{(D)}~183}</math>.
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
-{{AMC10 box|year=2024|ab=A|num-b=24|after=Last Problem}}
-[[Category:Intermediate Geometry Problems]]
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