Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 6.4 National olympiad Prove it Spain

Let $a$ , $b$ and $c$ be real numbers such that $p(x) = x^4 + a x^3 + b x^2 + a x + c$ has exactly three different real roots; these roots are $\tan y$ , $\tan 2y$ and $\tan 3y$ for some real number $y$ . Find all possible values of $y$ , $0 \le y < \pi$ .

Solution

that is,
$\tan(2ky) + \tan(my) + \tan(ny) = 0$
provided that $r$ , $s$ , $t$ are $\tan(ky)$ , $\tan(my)$ and $\tan(ny)$ respectively.
We consider now the following cases:
* If $r = \tan y$ , $s = \tan 2y$ and $t = \tan 3y$ then $\tan 2y + \tan 5y = 0$ , and
$y \in \left\{ \frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7}, \frac{4\pi}{7}, \frac{5\pi}{7}, \frac{6\pi}{7} \right\}.$
* If $r = \tan 2y$ , $s = \tan y$ and $t = \tan 3y$ then $\tan 4y + \tan 4y = 0$ and it follows that
$y \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \right\},$
We have discarded $y = \frac{\pi}{2}$ , $y = \frac{\pi}{4}$ and $y = \frac{3\pi}{4}$ because $\tan y$ and $\tan(2y)$ must be real numbers.
* If $r = \tan 3y$ , $s = \tan y$ and $t = \tan 2y$ then $\tan 6y + \tan 3y = 0$ and
$y \in \left\{ \frac{\pi}{9}, \frac{2\pi}{9}, \frac{\pi}{3}, \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{2\pi}{3}, \frac{7\pi}{9}, \frac{8\pi}{9} \right\}.$

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