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Given: $$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$

What is the sum of all arctan of all the roots of $p(x)$?

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2 Answers 2

up vote 17 down vote accepted

Let $p_1,p_2,p_3,p_4$ be the roots of $p(x)=0$

So, we need $\arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4$

Using Vieta's Formulae,

$\sum p_i=p_1+p_2+p_3+p_4=\frac{5773}1$

$\sum p_ip_j=p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4=\frac{46464}1$

$\sum p_ip_jp_k=p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4=\frac{5773}1$

$p_1p_2p_3p_4=\frac{46}1$

We know, $$\tan(A+B+C+D)$$

$$=\frac{\tan A+\tan B+\tan C+\tan D -(\tan A\tan B\tan C+\tan A\tan B\tan D+\tan A\tan C\tan D+\tan B\tan C\tan D)}{1- (\tan A\tan B+\tan A\tan C+\tan A\tan D+\tan B\tan C+\tan B\tan D+\tan C\tan D)+ \tan A\tan B\tan C\tan D}$$

If we put $p_1=\tan A,p_2=\tan B$ etc.,

$$\tan(A+B+C+D)=\frac{p_1+p_2+p_3+p_4 -(p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4)}{1- (p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4)+ p_1p_2p_3p_4}$$

$$=\frac{5773-5773}{1-46464+46}=0$$

So, $\tan(A+B+C+D)=0\implies A+B+C+D=\arctan (0)$

$\implies \arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4=\arctan (0)$

The general value of $\arctan (0)$ is $n\pi$ where $n$ is any integer, the special value being $0$(putting $n=0$)

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1  
i was really hoping for something a little more elementary, since I really can't understand any of these deductions (im senior high school student). I was hoping these olympiad questions were within reach. Thanks anyways! –  Or Cyngiser Dec 24 '12 at 19:57
2  
That's pretty much the simplest way to do it, really. –  Joe Z. Dec 24 '12 at 20:08
    
@JoeZeng, thanks. Wish there is a simpler way for Or Cyngiser. –  lab bhattacharjee Dec 24 '12 at 20:12
    
I think I see where the confusion is. You haven't expressed your answer in the form of arctans. Are A, B, C, and D the arctans of the solutions to p(x)? –  Joe Z. Dec 24 '12 at 20:19
    
@Joe Zeng: Yes the notation could be improved a bit. It is also not clear what $\sum \tan A$ and $\sum \tan A \tan B$ should stand for... (sum over what, if $A$ is the arctan of a solution...) –  Fabian Dec 24 '12 at 20:28

Nothing additional, but for the OP a more elementary approach.

$(x-a_1)(x-a_2)(x-a_3)(x-a_4) = x^4 - (a_1+a_2+a_3+a_4)x^3 + c_2x^2 - c_3x + a_1a_2a_3a_4.$ $c_2 = \Sigma_{i<j}a_ia_j$ and $c_3 = \Sigma_{i <j < k}a_ia_ja_k$. Let $t_i=\tan^{-1}a_i.$ You want to compute $\Sigma t_i.$ Here, the trick is knowing the $\tan()$ expansion $$\tan(x+y) = \frac{\tan x + \tan y}{1- \tan x \tan y}.$$
Apply recursively for $x=t_1+t_2$ and $y=t_3+t_4$. $$\tan(t_1+t_2+t_3+t_4) = \frac{\tan(t_1+t_2) + \tan(t_3+t_4)}{1+\tan(t_1+t_2)\tan(t_3+t_4)}=\frac{\frac{a_1 + a_2}{1- a_1a_2}+\frac{a_3 + a_4}{1- a_3a_4}}{1-\frac{a_1 + a_2}{1- a_1a_2}\frac{a_3 + a_4}{1- a_3a_4}} = \frac{(a_1+a_2)(1-a_3a_4)+(a_3+a_4)(1-a_1a_2)}{(1-a_1a_2)(1-a_3a_4)-(a_1+a_2)(a_3+a_4)}=\frac{c_1-c_3}{\dots}.$$ For this polynomial $c_1=c_3$, thefore $\tan\Sigma t_i =0,$ so is $\Sigma t_i =0$

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