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Difference between revisions of "2024 AMC 12A Problems/Problem 23"
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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>
 
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,  
+
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-\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>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>
 
</cmath>
−
Looking at the <math>x^4</math> term in the expansion for <math>f(x)</math> and using vietas gives
+
Looking at the <math>x^4</math> term in the expansion for <math>f(x)</math> and using vietas gives us
 
<cmath>
 
<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)  
 
\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)  

Revision as of 03:46, 9 November 2024

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,

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

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 Indentity)

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 4(Just do it)

Look at the options, A is too small and E is too big, you have BCD left, you can make a guess but can also estimate the values. After factorising in solution 1, tan(3pi/16)^2 is around 0.45,tan(5pi/16)^2 is around 2.24 and so on. Therefore, the answer is $\fbox{\textbf{(B) } 68}$.

Solution 5(transform)

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

See also

2024 AMC 12A (Problems • Answer Key • Resources)
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

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