Crazy Cosine Carousel

For $1 \le i \le 4$ , let $x_i = \cos t_i$ . Then the given equations become $2x_i^2 - 1 = 4x_i x_{i + 1},$ where the indices are taken modulo 4. Solving for $x_{i + 1}$ , we get $x_{i + 1} = \frac{2x_i^2 - 1}{4x_i}.$

Let $y_i = \frac{1}{x_i \sqrt{2}}$ , so $x_i = \frac{1}{y_i \sqrt{2}}$ . Substituting, we get $\frac{1}{y_{i + 1} \sqrt{2}} = \frac{2(\frac{1}{y_i \sqrt{2}})^2 - 1}{4 \cdot \frac{1}{y_i \sqrt{2}}},$ which simplifies to $y_{i + 1} = \frac{2y_i}{1 - y_i^2}.$

Finally, let $y_1 = \tan \theta$ , where $0^\circ \le \theta < 180^\circ$ . Then

$\begin{aligned} y_2 &= \frac{2y_1}{1 - y_1^2} = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \tan 2 \theta, \\ y_3 &= \frac{2y_2}{1 - y_2^2} = \frac{2 \tan 2 \theta}{1 - \tan^2 2 \theta} = \tan 4 \theta, \\ y_4 &= \frac{2y_3}{1 - y_3^2} = \frac{2 \tan 4 \theta}{1 - \tan^2 4 \theta} = \tan 8 \theta, \\ y_1 &= \frac{2y_4}{1 - y_4^2} = \frac{2 \tan 8 \theta}{1 - \tan^2 8 \theta} = \tan 16 \theta. \end{aligned}$

Hence, $\tan \theta = \tan 16 \theta$ . The $\tan$ function has period $180^\circ$ , so $15 \theta = (180n)^\circ$ for some integer $n$ , $0 \le n \le 14$ , or $\theta = (12n)^\circ.$ Then $(x_1, x_2, x_3, x_4)$ is of the form $\left( \frac{1}{\sqrt{2} \cdot \tan (12n)^\circ}, \frac{1}{\sqrt{2} \cdot \tan (24n)^\circ}, \frac{1}{\sqrt{2} \cdot \tan (48n)^\circ}, \frac{1}{\sqrt{2} \cdot \tan (96n)^\circ} \right).$

Using the identity $\frac{1}{\tan \theta} = \tan (90^\circ - \theta)$ , it is easy to check that up to cyclic permutation, all these solutions are equal to one of the following:

$\begin{array}{c} \left( \frac{1}{\sqrt{2}} \tan 30^\circ, -\frac{1}{\sqrt{2}} \tan 30^\circ, \frac{1}{\sqrt{2}} \tan 30^\circ, -\frac{1}{\sqrt{2}} \tan 30^\circ \right), \quad (1) \\ \left( \frac{1}{\sqrt{2}} \tan 18^\circ, -\frac{1}{\sqrt{2}} \tan 54^\circ, -\frac{1}{\sqrt{2}} \tan 18^\circ, \frac{1}{\sqrt{2}} \tan 54^\circ \right), \quad (2) \\ \left( \frac{1}{\sqrt{2}} \tan 6^\circ, -\frac{1}{\sqrt{2}} \tan 78^\circ, -\frac{1}{\sqrt{2}} \tan 66^\circ, -\frac{1}{\sqrt{2}} \tan 42^\circ \right), \quad (3) \\ \left( -\frac{1}{\sqrt{2}} \tan 6^\circ, \frac{1}{\sqrt{2}} \tan 78^\circ, \frac{1}{\sqrt{2}} \tan 66^\circ, \frac{1}{\sqrt{2}} \tan 42^\circ \right). \quad (4) \end{array}$

Because $x_i = \cos t_i$ , each element of a viable solution must be at most 1 in absolute value. Since $\frac{1}{\sqrt{2}} \tan 66^\circ > \frac{1}{\sqrt{2}} \tan 60^\circ = \frac{1}{\sqrt{2}} \cdot \sqrt{3} > 1,$ this rules out solutions (3) and (4).

On the other hand, $\tan 54^\circ = \frac{1 + \sqrt{5}}{\sqrt{10 - 2 \sqrt{5}}},$ and

$\begin{aligned} \frac{1}{\sqrt{2}} \cdot \frac{1 + \sqrt{5}}{\sqrt{10 - 2 \sqrt{5}}} &< 1 \\ \Leftrightarrow \quad 1 + \sqrt{5} &< 2 \sqrt{5 - \sqrt{5}} \\ \Leftrightarrow \quad (1 + \sqrt{5})^2 &< (2 \sqrt{5 - \sqrt{5}})^2 \\ \Leftrightarrow \quad 6 + 2 \sqrt{5} &< 20 - 4 \sqrt{5} \\ \Leftrightarrow \quad 6 \sqrt{5} &< 14 \\ \Leftrightarrow \quad 3 \sqrt{5} &< 7 \\ \Leftrightarrow \quad (3 \sqrt{5})^2 &< 7^2 \\ \Leftrightarrow \quad 45 &< 49, \end{aligned}$

which is true, so solutions (1) and (2) are viable.

The last step is to determine how many carousels can be made from solutions (1) and (2).

In solution (1), note that $x_1 = x_3$ and $x_2 = x_4$ . But all the $t_i$ must be distinct, so there is exactly one carousel that corresponds to solution (1), which is as follows: Let $\alpha = \arccos \left( \frac{1}{\sqrt{2}} \tan 30^\circ \right) = \arccos \left( \frac{1}{\sqrt{6}} \right).$ Then $0 < \alpha < 180^\circ$ and $\cos (180^\circ - \alpha) = -\cos \alpha = -\frac{1}{\sqrt{6}}$ , so $(\alpha, 180^\circ - \alpha, 360^\circ - \alpha, \alpha + 180^\circ)$ is a carousel.

In solution (2), all the $x_i$ are distinct. This means for each $i$ , there are two choices for $t_i$ , namey $t_i = \arccos x_i$ and $t_i = 360^\circ - \arccos x_i$ . So there are $2^4 = 16$ carousels that correspond to solution (2).

This gives us a total of $1 + 16 = 17$ carousels.

Let $u_j = \cos t_j$ for $1 \le j \le 4$ . Then $-1 \le u_j < 1$ for all $j$ , and $\begin{array}{ccc}2u_1^2-1 = 4u_1u_2 &\qquad& 2u_2^2-1 = 4u_2u_3 \\ 2u_3^2-1 = 4u_3u_4 &\qquad& 2u_4^2-1 = 4u_4u_1 \end{array}$ Thus $u_j \neq 0$ for all $j$ , and $u_2 \; = \; f(u_1) \qquad u_3 \; = \; f(u_2) \qquad u_4 \; = \; f(u_3) \qquad u_1 \; = \; f(u_4)$ where $f$ is the function $f(x) \; = \; \frac{2x^2-1}{4x}$ and hence, composing $f$ with itself $ffff(u_j) \; = \; u_j \qquad 1 \le j \le 4$ Now $ffff(x)-x$ is the rational function with numerator $\begin{array}{l} 1 - 208 x^2 + 5040 x^4 - 29120 x^6 + 22880 x^8 \\ + 109824 x^{10} - 163072 x^{12} + 56320 x^{14} - 3840 x^{16} \end{array}$ and denominator $\begin{array}{l} 32 x (-1 + 2 x^2) (1 - 12 x^2 + 4 x^4) \\ \times (1 - 56 x^2 + 280 x^4 - 224 x^6 + 16 x^8)\end{array}$ The numerator factorises as $\begin{array}{l} -(1 + 2 x^2) (-1 + 6 x^2) (1 - 20 x^2 + 20 x^4) \\ \times (1 - 184 x^2 + 536 x^4 - 224 x^6 + 16 x^8)\end{array}$ so we deduce that $u_1,u_2,u_3,u_4$ satisfy $h(x)=0$ , where $h(x)$ is the polynomial $(6 x^2-1) (1 - 20 x^2 + 20 x^4) (1 - 184 x^2 + 536 x^4 - 224 x^6 + 16 x^8)$ Thus the only possible values of $u_1,u_2,u_3,u_4$ are $\pm\alpha,\pm\beta_1,\pm\beta_2,\pm\gamma_1,\pm\gamma_2,\pm\gamma_3,\pm\gamma_4$ , where $\begin{array}{cc} \alpha=\frac{1}{\sqrt{6}} \\ \beta_1=\sqrt{\frac{5+2\sqrt{5}}{10}} & \beta_2=\sqrt{\frac{5 - 2\sqrt{5}}{10}} \\ \gamma_1=\sqrt{\tfrac72 - \sqrt{5} - \sqrt{3(5-2\sqrt{5})}} & \gamma_2=\sqrt{\tfrac72 - \sqrt{5} + \sqrt{3(5-2\sqrt{5})}} \\ \gamma_3=\sqrt{\tfrac72 + \sqrt{5} - \sqrt{3(5+2\sqrt{5})}} & \gamma_4=\sqrt{\tfrac72 + \sqrt{5} + \sqrt{3(5+2\sqrt{5})}} \end{array}$ But $f(\gamma_1) = -\gamma_4 \qquad f(\gamma_2) = \gamma_3 \qquad f(\gamma_3) = -\gamma_1 \qquad f(\gamma_4) = \gamma_2$ and so the only possible candidates for $u$ -translates of carousels involving the $\gamma$ terms are cyclic permutations of $\gamma_1,-\gamma_4,-\gamma_2,-\gamma_3 \qquad \mbox{and} \quad -\gamma_1,\gamma_4,\gamma_2,\gamma_3$ Since $\gamma_2,\gamma_4 > 1$ , we deduce that none of $\pm\gamma_1,\pm\gamma_2,\pm\gamma_3,\pm\gamma_4$ can occur in a $u$ -translate of a carousel. Thus the only possible values of $u_1,u_2,u_3,u_4$ are $\pm\alpha,\pm\beta_1,\pm\beta_2$ , and since $f(\pm\alpha) \; =\; \mp\alpha \qquad f(\pm\beta_1) \; = \;\pm\beta_2 \qquad f(\pm\beta_2) \; = \; \mp\beta_1$ the only possible values of $u_1,u_2,u_3,u_4$ are $\alpha,-\alpha,\alpha,-\alpha \qquad \mbox{or} \qquad \beta_1,\beta_2,-\beta_1,-\beta_2$ or some cyclic permutation of one of these.

We need to convert these two $u$ -translates back into the possible carousels. For any $0<|u|<1$ there are exactly $2$ values of $t$ between $0$ and $2\pi$ such that $\cos t = u$ .

1. Since the four numbers in a carousel need to be distinct, there is exactly one carousel which leads to the $u$ -translate $\alpha,-\alpha,\alpha,-\alpha$ , and that is $\{\cos^{-1}\alpha,\pi-\cos^{-1}\alpha,2\pi-\cos^{-1}\alpha,\pi+\cos^{-1}\alpha \}$ .

2. To get the $u$ -translate $\beta_1,\beta_2,-\beta_1,-\beta_2$ , we need to choose one out of $\cos^{-1}\beta_1$ and $2\pi-\cos^{-1}\beta_1$ , one out of $\cos^{-1}\beta_2$ and $2\pi-\cos^{-1}\beta_2$ , one out of $\pi - \cos^{-1}\beta_1$ and $\pi + \cos^{-1}\beta_1$ , and one out of $\pi - \cos^{-1}\beta_2$ and $\pi + \cos^{-1}\beta_2$ . Since the eight numbers we are choosing between are all distinct, we can make all four of these choices freely and always obtain four distinct numbers, and hence there are $16$ different carousels which lead to the $u$ -translate $\beta_1,\beta_2,-\beta_1,-\beta_2$ .

Thus there are $17$ carousels in total.