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

Revision as of 20:17, 23 December 2024

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(single formula)
8 Solution 8(just do it ✅）
9 Solution 9 (Vietas)
10 Alternate proof of the two tangent squares formula
11 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,

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

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 $(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(4)=60$ $(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}$

@@ Line 212: / Line 212: @@
 <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==
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=22|num-a=24}}
 {{MAA Notice}}

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"

Revision as of 20:17, 23 December 2024

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(single formula)

Solution 8(just do it ✅）

Solution 9 (Vietas)

Alternate proof of the two tangent squares formula

See also