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ganeshie8
 one year ago
find a third degree polynomial with integer coefficients whose roots are \[\cot^2\frac{\pi}{7},~\cot^2\frac{2\pi}{7},~\cot^2\frac{3\pi}{7}\]
ganeshie8
 one year ago
find a third degree polynomial with integer coefficients whose roots are \[\cot^2\frac{\pi}{7},~\cot^2\frac{2\pi}{7},~\cot^2\frac{3\pi}{7}\]

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the sum and product of roots do give rational numbers http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5E3+%28cot%28k*pi%2F7%29%29%5E2 http://www.wolframalpha.com/input/?i=%5Cprod%5Climits_%7Bk%3D1%7D%5E3+%28cot%28k*pi%2F7%29%29%5E2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider the polynomial equation: \(x^6 +x^5 + x^4 + x^3 + x^2 + x + 1 = 0\) It's roots are \(e^{i2k\pi/7}\) where \(k = 1\) to \(6\) Divide the equation by \(x^3\) , then we have, \(x^3 + x^2 + x + 1 + 1/x + 1/x^2 + 1/x^3 = 0 \) \( (x+1/x)^3 3(x+1/x) + (x+1/x)^2  2 + (x+1/x) + 1 = 0 \) When x = \(e^{i2k\pi/7}\) , \((x+1/x) = 2\cos(2k\pi/7)\) So, \(y^3 + y^2 2y 1 = 0\) has the roots \(2\cos(2\pi/7), 2\cos(4\pi/7) , 2\cos(6\pi/7)\) Now, note that \( \cos(2k\pi/7) = \cos^2(k\pi/7)  \sin^2(k\pi/7)\) \(= \frac{1}{\sec^2(k\pi/7)}  \frac{\tan^2(k\pi/7)}{\sec^2(k\pi/7)}\) \(= \frac{1\tan^2(k\pi/7)}{1 + \tan^2(k\pi/7)} \) \(= \frac{\cot^2(k\pi/7)  1}{\cot^2(k \pi/7) + 1}\) So, if \(\cot^2(k\pi/7) = t\) , \(2\cos(2k\pi/7) = 2\frac{(t1)}{(t+1)}\) Therefore \(\cot^2(k\pi/7)\) satisfies the following equation \( (2\frac{(t1)}{(t+1)})^3 + (2\frac{(t1)}{(t+1)})^2 2(2\frac{(t1)}{(t+1)})  1 = 0\) Multiplying both sides by \((t+1)^3\) , \( (2t2)^3 + (t+1)(2t2)^2 2(t+1)^2(2t2)  (t+1)^3 = 0\) Therefore, \( (2t2)^3 + (t+1)(2t2)^2 2(t+1)^2(2t2)  (t+1)^3 \) is clearly a polynomial with integer coefficients and has as its roots \(\cot^2(k\pi/7)\) for \(k = 1,2,3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also, simplifying the final expression, the final polynomial turns out to be \(7t^3  35t^2 + 21t1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2beautiful! it seems there is no question i know which you cannot solve !

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0Andre thats make me cry :\ @adxpoi lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, nah I'm sure there's stuff I can't solve :P

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0extremely interesting question hmmm and answer :O

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Hey I'm pretty sure I've asked this one before

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think @adxpoi 's method generalizes to any arbitrary degree polynomial : Find a polynomial of \(n\)th degree whose roots are \[\cot^2\frac{\pi}{2n+1},~\cot^2\frac{2\pi}{2n+1},~\cot^2\frac{3\pi}{2n+1},\cdots, \cot^2\frac{n\pi}{2n+1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes @ParthKohli I remember, we had solved it using Vieta's formulas or something adhoc... :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1No, complex numbers.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1In fact just like how adx has solved it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ahh okay, can't tryst my memory lol knw how to solve the general case for arbitrary degree ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1My book (KD Joshi) has this, yeah.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ohk.. I'm hoping we can simply mimic the adx's method, but I haven't tried it yet... so not completely sure yet..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Where did you find this question?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2IMO shortlisted practice problems

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2@praxer has been asking these recently

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, this one is pretty standard for an IMO problem if you've read about it.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I read and forget most of what I read, so I must have encountered this problem before.. but I'm pretty sure I never had this observation before : \[x^6+x^5+\cdots+x+1 = \dfrac{x^71}{x1}\] so \(e^{i2k\pi/7}\) is a root

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Is there any other way to see that \(e^{i2k\pi/7}\) satisfies that polynomial equation ? (w/o using geometric series)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Here's a directdownload link to the book I'm talking about: https://doc0o44docs.googleusercontent.com/docs/securesc/ha0ro937gcuc7l7deffksulhg5h7mbp1/l4j8b8u0i1870es3h7u201vqlntqv62q/1439287200000/06023952678462293309/*/0BwEaIp7y7BJTOVZVcVZiVUw3aXM?e=download The author explains the general case really nicely in the trigonometric identities or equations chapter (I don't remember which one it was).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Pages 246247, yes.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2slightly different problem, but the methods are similar yeah https://i.gyazo.com/27c20454519da07d956f53137adc3ee6.png
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