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

Revision as of 10:27, 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)

There are five options for this question, 28,68,70,72,84, respectively. Since 28 is too small and 84 is too large, these two options may be eliminated. Now, only three options are left. At this point, you can take a guess if you want to. However,we should take a closer look at these options. When dividing these options by two, we obtain 34,35,and 36 respectively. Option C should be eliminated because it is the only odd number among these three options when divided by two. Last but not least, Option D, which is 72, appears too much in AMC choices. Therefore, I guess that the MAA will try not use this option again for AMC, eliminating D. As a result, the only option remaining is B.

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"

Revision as of 10:27, 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 247: / Line 247: @@
 ~ilikemath247365
+==Solution 10 (Options)==
+There are five options for this question, 28,68,70,72,84, respectively. Since 28 is too small and 84 is too large, these two options may be eliminated. Now, only three options are left. At this point, you can take a guess if you want to. However,we should take a closer look at these options. When dividing these options by two, we obtain 34,35,and 36 respectively. Option C should be eliminated because it is the only odd number among these three options when divided by two. Last but not least, Option D, which is 72, appears too much in AMC choices. Therefore, I guess that the MAA will try not use this option again for AMC, eliminating D. As a result, the only option remaining is B.