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Difference between revisions of "2024 AMC 12A Problems/Problem 23"
(→‎Alternate proof of the two tangent squares formula)
 
(40 intermediate revisions by 13 users not shown)
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First, notice that
 
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} \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{4\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>
  
  
Line 19: Line 19:
 
<cmath>=(\tan x+\tan (\frac{\pi}{2}-x))^2-2</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 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 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{\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{1}{\cos x \sin x}\right)^2-2</cmath>
 
<cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath>
 
<cmath>=\left(\frac{2}{\sin 2x}\right)^2-2</cmath>
Line 27: Line 27:
 
Hence,
 
Hence,
  
−
<cmath>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})*(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})</cmath>
+
<cmath>(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})</cmath>
−
<cmath>=(\frac{4}{\sin^2 \frac{\pi}{8}}-2)(\frac{4}{\sin^2 \frac{3\pi}{8}}-2)</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  
 
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{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}</cmath>
 +
  
−
and
+
<cmath>\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\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,
 
Hence,
  
−
<cmath>(\frac{4}{\sin^2 \frac{\pi}{8}}-2)(\frac{4}{\sin^2 \frac{3\pi}{8}}-2)</cmath>
+
<cmath>\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)</cmath>
  
−
<cmath>=(\frac{16}{2-\sqrt{2}}-2)(\frac{16}{2+\sqrt{2}}-2)</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>=(14+8\sqrt{2})(14-8\sqrt{2})</cmath>
Line 48: Line 47:
 
<cmath>=68</cmath>
 
<cmath>=68</cmath>
  
−
Therefore, the answer is \fbox{\textbf{(B) } 68}.
+
Therefore, the answer is <math>\fbox{\textbf{(B) } 68}</math>.
  
 
~tsun26
 
~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>
 +
 +
 +
<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(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 8(just do it ✅）==
 +
Since A is too big and E is too small, There is only 3 options left, you can make a guess now, however, estimating the value of it is also okay. You will get 68 for the final answer.
 +
 +
==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
 +
 +
<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>
 +
<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>.
 +
 +
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>
 +
 +
==Alternate proof of the two tangent squares formula==
 +
 +
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>. Notice we can write:
 +
<math>(tanx + cotx)^{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}}{tanx}</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}{sinxcosx}</math>. So, we have that: <math>\tan^{2}(x)</math> + <math>\cot^{2}(x)</math> = <math>\frac{1}{sinxcosx}^{2}</math> - <math>{2}</math> which can be simplfied to: 2(<math>\frac{2}{sin^{2}(2x)}</math> - 1) or (<math>\frac{4}{sin^{2}(2x)}</math> - 2) as stated in earlier solutions.
 +
 +
~ilikemath247365
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:32, 23 December 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 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(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 8(just do it ✅）

Since A is too big and E is too small, There is only 3 options left, you can make a guess now, however, estimating the value of it is also okay. You will get 68 for the final answer.

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: $(tanx + cotx)^{2}$ 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 $\frac{\sec^{2}{x}}{tanx}$. 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 $\frac{1}{sinxcosx}$. So, we have that: $\tan^{2}(x)$ + $\cot^{2}(x)$ = $\frac{1}{sinxcosx}^{2}$ - ${2}$ which can be simplfied to: 2($\frac{2}{sin^{2}(2x)}$ - 1) or ($\frac{4}{sin^{2}(2x)}$ - 2) as stated in earlier solutions.

~ilikemath247365

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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