Difference between revisions of "2024 AMC 12A Problems/Problem 23"
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~tsun26 | ~tsun26 | ||
| − | ==Solution 2 (Another | + | ==Solution 2 (Another Identity)== |
First, notice that | First, notice that | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:KEVIN_LIU KEVIN_LIU] | ~[https://artofproblemsolving.com/wiki/index.php/User:KEVIN_LIU KEVIN_LIU] | ||
| − | ==Solution | + | ==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> | 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> | ||
| Line 153: | Line 150: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
| + | |||
==Solution 6 (Half angle formula twice)== | ==Solution 6 (Half angle formula twice)== | ||
So from the question we have: | So from the question we have: | ||
| Line 182: | Line 180: | ||
~ERiccc | ~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
Contents
- 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 ![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
Solution 1 (Trigonometric Identities)
First, notice that
![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}\]](http://latex.artofproblemsolving.com/a/e/f/aef6608848ffff3cf92d16b38e25d9445192ffa8.png)
![\[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\]](http://latex.artofproblemsolving.com/6/8/b/68be2a9a09e416cd1b012079639d6edec9573d00.png)
Here, we make use of the fact that
![\[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\]](http://latex.artofproblemsolving.com/d/9/c/d9cc31bca74f57bf2838b0b6598bc1f00dff7571.png)
![\[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\]](http://latex.artofproblemsolving.com/b/6/7/b6793d18f913669880bf136ac07b85de29437588.png)
![\[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/7/9/3/7933b5b30ffa9897d781738ab273c23ef1bb28c8.png)
![\[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/1/b/2/1b2030fb90cbf1e3c35d71c2058af8d5ba22dff7.png)
![\[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]](http://latex.artofproblemsolving.com/c/f/7/cf70707a0104f210582ad6d9d6c6102e9d9c65a3.png)
![\[=\left(\frac{1}{\cos x \sin x}\right)^2-2\]](http://latex.artofproblemsolving.com/7/0/9/709c6b08ef0737e8b6bea68c03278d9c9c9c1f5b.png)
![\[=\left(\frac{2}{\sin 2x}\right)^2-2\]](http://latex.artofproblemsolving.com/6/7/a/67a0346eedaba73caf1bd1565dade45e9f18032f.png)
![\[=\frac{4}{\sin^2 2x}-2\]](http://latex.artofproblemsolving.com/6/6/c/66cf0347f5720950827b3ccc7539a5cd1ff39a18.png)
Hence,
![\[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\]](http://latex.artofproblemsolving.com/3/7/0/37058506cfc079d9fda277c9f5204d8fdd24397a.png)
![\[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]](http://latex.artofproblemsolving.com/d/e/b/debb6db041d8ae12049f00e2a1e502bcdf1fd554.png)
Note that
![\[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\]](http://latex.artofproblemsolving.com/c/0/d/c0d937d7c96c598596abf879a2858dfe03279c78.png)
![\[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\]](http://latex.artofproblemsolving.com/a/9/0/a90264ad7a0531c9c468da3648ae7f189b84c632.png)
Hence,
![\[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]](http://latex.artofproblemsolving.com/c/6/4/c64ea214245a92d0b7e7ed3ec29d9a7b9a4f4b4e.png)
![\[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\]](http://latex.artofproblemsolving.com/0/4/4/044f7e8a8a39685da59a37f351e39327160c09b5.png)
![\[=(14+8\sqrt{2})(14-8\sqrt{2})\]](http://latex.artofproblemsolving.com/f/2/c/f2c6a5cb2a1fb1fde4bb562119d2e31685b8a08e.png)
![\[=68\]](http://latex.artofproblemsolving.com/5/a/d/5ad6af812b7642452ebc0775651efbcd868b54aa.png)
Therefore, the answer is 
.
~tsun26
Solution 2 (Another Identity)
First, notice that
Here, we make use of the fact that
Hence,
Therefore, the answer is .
Solution 3 (Complex Numbers)
Let 
. Then,
![\[y = e^{8i\theta} = e^{\frac{\pi}{2} i} = (\cos \theta + i\sin \theta)^8 = 0 + i.\]](http://latex.artofproblemsolving.com/4/8/1/48163fb06cd77ea3dd09c1da9c1d593e67615ac5.png)
Expanding by using a binomial expansion,
![\[\Re(y) = \cos^8 \theta - 28 \cos^6 \theta \sin^2 \theta + 70 \cos^4 \theta \sin^4 \theta - 28 \cos^2 \theta \sin^6 \theta + \sin^8\theta =0.\]](http://latex.artofproblemsolving.com/d/6/f/d6f804da8af7cc8eebe12a27568553abd4aab76a.png)
Divide by 
and notice we can set 
where 
. Then, define 
so that
![\[f(x) = 1 - 28 x^2 + 70 x^4 - 28 x^6 + x^8.\]](http://latex.artofproblemsolving.com/f/4/b/f4b25507d3cbfc37aba70da6054414659b1f4c35.png)
Notice that we can have 
because we are only considering the real parts. We only have this when 
, meaning 
. This means that we have 
as unique roots (we get them from ![$k\theta \in [0,\pi]$](http://latex.artofproblemsolving.com/e/f/5/ef55e16a7aaae7add05f03cdc1afb5773a37ade0.png)
) and by using the fact that 
, we get ![\[x \in \left\{\tan \theta, -\tan \theta, \tan \left(3 \theta \right), -\tan \left(3 \theta \right), \tan \left(5 \theta \right), -\tan \left(5 \theta \right), \tan \left(7 \theta \right), -\tan \left(7 \theta \right) \right\}\]](http://latex.artofproblemsolving.com/2/7/2/272510786c7e81c31e734c70b1fa2cf0a6c7183d.png)
Since we have a monic polynomial, by the Fundamental Theorem of Algebra,
![\[f(x) = (x-\tan \theta)(x+\tan \theta) (x-\tan \left(3 \theta \right))(x+\tan \left(3 \theta \right)) (x-\tan \left(5 \theta \right))(x+\tan \left(5 \theta \right))(x-\tan\left(7 \theta \right))(x+\tan \left(7 \theta \right))\]](http://latex.artofproblemsolving.com/9/a/b/9abc9f2eb9ee6db6b9b4ad3b04612a164e2056e0.png)
![\[f(x) = (x^2 - \tan^2 \theta)(x^2 - \tan^2 (3\theta))(x^2 - \tan^2 (5\theta))(x^2 - \tan^2 (7\theta))\]](http://latex.artofproblemsolving.com/e/8/2/e82d837dbfb2368d1659a70f0c50e663a5616241.png)
Looking at the 
term in the expansion for and using vietas gives us
![\[\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 \theta \tan^2 (7\theta) + \tan^2 (3\theta) \tan^2 (5\theta)\]](http://latex.artofproblemsolving.com/4/0/4/4043f9c5c4aaa5b675fbf0c62c5349ae4b2ee844.png)
![\[+ \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) = \frac{70}{1} = 70.\]](http://latex.artofproblemsolving.com/7/3/b/73bcc59b9761e8ffd8ec29ccf1e7ed7805ff9ca5.png)
Since 
and 
![\[\tan^2 \theta \tan^2 (7\theta) = \tan^2 (3\theta) \tan^2 (5\theta) = 1.\]](http://latex.artofproblemsolving.com/4/7/1/47178c719a5a0cbebbbaf85821b004fd16f16aeb.png)
Therefore
![\[\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) + 2 = 70.\]](http://latex.artofproblemsolving.com/0/7/8/0783858284b4fb251b62020f2b4e94a8f59d2b62.png)
![\[\tan^2 \theta \tan^2 (3\theta) + \tan^2 \theta \tan^2 (5\theta) + \tan^2 (3\theta) \tan^2 (7\theta) + \tan^2 (5\theta) \tan^2 (7\theta) = \boxed{\textbf{(B) } 68}\]](http://latex.artofproblemsolving.com/a/e/1/ae158c94b859d59983493b715b0d0d01601bdb4b.png)
Solution 5 (Transformation)
Set x = 
, 7x = 
- x ,
set C7 = 
, C5 = 
, C3 = 
, C= 
, S2 = 
, S6 = 
First, notice that
![\[\tan^2 x \cdot \tan^2 3x + \tan^2 3x \cdot \tan^2 5x+\tan^2 3x \cdot \tan^2 7x+\tan^2 5x \cdot \tan^2 7x\]](http://latex.artofproblemsolving.com/d/0/0/d0069bbe4a2616e99e7ab0491e3a64d2000680ca.png)
![\[=(\tan^2x+\tan^2 7x)(\tan^23x+\tan^2 5x)\]](http://latex.artofproblemsolving.com/c/8/6/c863068785cfd435b5966cde9824360f2661d8f0.png)
![\[=(\frac{1}{C} - 1 +\frac{1}{C7}-1)(\frac{1}{C3} - 1 +\frac{1}{C5}-1)\]](http://latex.artofproblemsolving.com/5/e/c/5ece63b9745d50725594be0ba20c801af85ce7da.png)
![\[=(\frac{C+C7}{C \cdot C7} -2)( \frac{C3+C5}{C3 \cdot C5} -2)\]](http://latex.artofproblemsolving.com/5/7/4/574291dc818f9e01449abd78c40431dd34d70ad3.png)
![\[=(\frac{1}{C \cdot S} -2)( \frac{1}{C3 \cdot S3} -2)\]](http://latex.artofproblemsolving.com/6/3/3/633de8a32edcf16ff3afb4841ad330f1bbf0fb2f.png)
![\[=(\frac{4}{S2} -2)( \frac{4}{S6} -2)\]](http://latex.artofproblemsolving.com/4/f/b/4fb736f04b922430a702441a798072135d527c10.png)
![\[=4(\frac{2-S2}{S2})( \frac{2-S6}{S6})\]](http://latex.artofproblemsolving.com/4/e/4/4e481b8fc6bc04cb3fcfcb8ab5ddc45092753973.png)
![\[=4(\frac{4-2 \cdot S2-S \cdot S6 }{S2 \cdot S6}+1)\]](http://latex.artofproblemsolving.com/5/3/1/53175c2b8e3ab4976484e6667b4a2be92b39be14.png)
![\[=4 + \frac{8}{S2 \cdot S6}\]](http://latex.artofproblemsolving.com/c/4/5/c455b9a45428963582dc614a7c3e9b8fa3ad24d0.png)
![\[=4 + \frac{32}{S4}\]](http://latex.artofproblemsolving.com/c/6/3/c63e90fa66d90ab74315f9040a22d15acec5ac18.png)
![\[=4 + 64\]](http://latex.artofproblemsolving.com/9/1/7/91707b16bb351f37f9bcffe668390f25c4e67e2b.png)
![\[= 68\]](http://latex.artofproblemsolving.com/a/3/b/a3b1f4ddc39ee015eb6f4edb6e31c1605215d208.png)
Solution 6 (Half angle formula twice)
So from the question we have:
Using 
![\[=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1+\cos\frac{7\pi}{8}}{1-\cos\frac{7\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1+\cos\frac{5\pi}{8}}{1-\cos\frac{5\pi}{8}})\]](http://latex.artofproblemsolving.com/5/6/e/56ebe722b6e2f419cd86b604d8350d593bb12190.png)
Using 
![\[=(\frac{1+\cos\frac{\pi}{8}}{1-\cos\frac{\pi}{8}}+\frac{1-\cos\frac{\pi}{8}}{1+\cos\frac{\pi}{8}})(\frac{1+\cos\frac{3\pi}{8}}{1-\cos\frac{3\pi}{8}}+\frac{1-\cos\frac{3\pi}{8}}{1+\cos\frac{3\pi}{8}})\]](http://latex.artofproblemsolving.com/7/0/e/70eddc802a8b255cc2493fb586b9bbecf80177ae.png)
![\[=(\frac{(1+\cos\frac{\pi}{8})^2+(1-\cos\frac{\pi}{8})^2}{1-\cos^2\frac{\pi}{8}})(\frac{(1+\cos\frac{3\pi}{8})^2+(1-\cos\frac{3\pi}{8})^2}{1-\cos^2\frac{3\pi}{8}})\]](http://latex.artofproblemsolving.com/f/2/3/f238fd5aea2b64660f4aabfcfc938dfdedc94c57.png)
![\[=(\frac{2+2\cos^2\frac{\pi}{8}}{1-\cos^2\frac{\pi}{8}})(\frac{2+2\cos^2\frac{3\pi}{8}}{1-\cos^2\frac{3\pi}{8}})\]](http://latex.artofproblemsolving.com/d/7/0/d70c116fd21a90b623719188b93925343dca3e7e.png)
Using 
![\[=(\frac{2+1+\cos\frac{\pi}{4}}{1-\frac{1+\cos\frac{\pi}{4}}{2}})(\frac{2+1+\cos\frac{3\pi}{4}}{1-\frac{1+\cos\frac{3\pi}{4}}{2}})\]](http://latex.artofproblemsolving.com/5/b/5/5b5bfc71af3aab6216b25172f9b70106f6fc58fe.png)
![\[=(\frac{12+2\sqrt{2}}{4-2\sqrt{2}})(\frac{12-2\sqrt{2}}{4+2\sqrt{2}})\]](http://latex.artofproblemsolving.com/5/e/b/5eb17d9a350db96240c839962a8a1648041f6329.png)
![\[=\frac{136}{2}=\boxed{\textbf{B) }68 }\]](http://latex.artofproblemsolving.com/9/d/f/9dfaeadcd4f3c2bc954482a7ae3745ccff28b75d.png)
~ERiccc
Solution 7(single formula)
![\[\cot \alpha - \tan \alpha = 2 \cot 2 \alpha \implies \cot^2 \alpha + \tan^2 \alpha = 4 \cot^2 2 \alpha + 2.\]](http://latex.artofproblemsolving.com/c/a/9/ca93ddf4e187e5f1981bd422a3dd638e73b67a44.png)
We use 
for 
![\[(\tan^2 \alpha + \cot^2 \alpha)(\tan^2 (\frac{\pi}{4} - \alpha) + \cot^2 (\frac{\pi}{4} - \alpha)) = (4 \cot^2 2 \alpha + 2)(4 \cot^2 (\frac{\pi}{2} - 2\alpha) +2) =\]](http://latex.artofproblemsolving.com/2/3/8/238726747d82f2ea7ca17e1a6730b22786835b20.png)
![\[= 4 \cdot(4+ 2\tan^2 2\alpha + 2\cot^2 2\alpha +1) = 20 + 8 \cdot (4 \cot^2 4 \alpha +2) = 68.\blacksquare\]](http://latex.artofproblemsolving.com/7/4/2/74225e6a59c7b79c511bc294d4947be125b2019b.png)
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 
.
Before we directly solve this problem, let's analyze the roots of 
, or equivalently using tangent expansion formula, 
, which implies 
. Now note that the roots of this equation are precisely 
, so the second symmetric sum of these four numbers is 
by Vieta's. Thus, we have ![\[\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6\]](http://latex.artofproblemsolving.com/7/7/f/77f68c5cfaa622e39e395b39e7b7adbfbac6abe1.png)
Upon further inspection, 
using the fact that 
. Hence, we have ![\[\tan\frac{\pi}{16}\tan\frac{5\pi}{16}-1+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}-1+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=6\]](http://latex.artofproblemsolving.com/9/8/b/98b6507c6e60cf120dc221337bda7e45a628437c.png)
![\[\tan\frac{\pi}{16}\tan\frac{5\pi}{16}+\tan\frac{\pi}{16}\tan\frac{13\pi}{16}+\tan\frac{5\pi}{16}\tan\frac{9\pi}{16}+\tan\frac{9\pi}{16}\tan\frac{13\pi}{16}=8\]](http://latex.artofproblemsolving.com/d/8/e/d8ef1d4e1970b2ee0d64c16a061790308dff33b5.png)
![\[(\tan\frac{\pi}{16}+\tan\frac{9\pi}{16})(\tan\frac{5\pi}{16}+\tan\frac{13\pi}{16})=8\]](http://latex.artofproblemsolving.com/5/f/5/5f580077326b71274c4cb1290ffe82d66dcdb2df.png)
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
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16}-2)(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16}-2)=64\]](http://latex.artofproblemsolving.com/6/c/8/6c8b8bba94d3ce1b69bef5c76f120a6871be9f0d.png)
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})+4=64\]](http://latex.artofproblemsolving.com/4/e/b/4eb3fa9c9e4b7bc76a6c84ca8fdd5efa4f8e7127.png)
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{9\pi}{16})(\tan^2\frac{5\pi}{16}+\tan^2\frac{13\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60\]](http://latex.artofproblemsolving.com/2/6/c/26cf36082d0089aa5e038f3dca66b106b80ba4ca.png)
Then, use the fact that 
to get
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(\tan^2\frac{\pi}{16}+\tan^2\frac{5\pi}{16}+\tan^2\frac{9\pi}{16}+\tan^2\frac{13\pi}{16})=60\]](http://latex.artofproblemsolving.com/e/2/f/e2f365e276cdbf588f41c3d98462cd1c5994b399.png)
Hold on; the first term is exactly what we are solving for! It thus suffices to find 
. Fortunately, this is just 
(Where 
is the nth symmetric sum), with relation to roots of . By Vieta's, this is just 
.
Finally, we plug this value into our equation to obtain
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})-2(4)=60\]](http://latex.artofproblemsolving.com/1/4/5/145374bce3df5dd3a333c3a5c1576a57e6998d0e.png)
![\[(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16})=\boxed{68}\]](http://latex.artofproblemsolving.com/3/c/9/3c989559b09df69eb8e2f5e4373db97d4ff66236.png)
Alternate proof of the two tangent squares formula
We want to simplify 
+ 
. We make use of the fact that 
= 
. Then, the expression becomes + 
. Notice we can write:
as + + 2 as tangent and cotangent are reciprocals of each other. Then, the sum of the tangent and cotangent can be simplified to 
. 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 
. So, we have that: + = 
- 
which can be simplfied to: 2(
- 1) or (
- 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
