Dynamical System

Start from the element $a$ . Put $b=f(a)$ . Note that the equation $a=f(a)$ has no real solutions, so $b\neq a$ .

$f(f(a))=f(b)=\frac{1+\frac{1+a}{1-a}}{1-\frac{1+a}{1-a}}=\frac{1-a+1+a}{1-a-(1+a)}=-\frac1a$

So $f(f(f(f(a))))=a$ . There are four elements generated from $a$ :

$a,b,-\frac1a,-\frac1b$

and their product is $1$ . The set $A$ is entirely made up of these cycles, so the overall product is $\boxed1$ .

The final point to note is that none of the numbers $-1,0,1$ can be in $A$ , as the fourth element of their cycle is $\infty$ .

Recall the angle addition formula for tangent, and consider the result of setting $\theta = 45^\circ$ so that $\tan \theta = 1$ . $\tan (\theta + \alpha) = \frac{\tan \theta + \tan \alpha}{1 - \tan \theta \tan \alpha}$ $\tan (45^\circ + \alpha) = \frac{1 + \tan \alpha}{1 - \tan \alpha}$

This formula has a very similar structure to the function $f(a)$ in this problem, so let's try for a geometric interpretation where $a$ and $f(a)$ are slopes in the xy -plane. If we identify $a = \tan \alpha$ , then $f(a) = \tan (45^\circ + \alpha)$ . So in the xy -plane, if $a$ is the slope of a given line, then $f(a)$ is the slope of a second line rotated $45^\circ$ counterclockwise from the first . Continuing to apply $f$ simply rotates the line an additional $45^\circ$ and reports the new slope.

Regardless of the starting slope $a$ , four applications of $f$ will have rotated the line $180^\circ$ , giving us back the original line and slope. Since set $A$ is closed under application of $f$ , we see that the entries in $A$ must come in sets of four: the slopes of four lines at angles $\alpha$ , $\alpha + 45^\circ$ , $\alpha + 90^\circ$ and $\alpha + 135^\circ$ . The first and third lines are perpendicular with opposite reciprocal slopes, as are the second and fourth lines. So the product of the four slopes will always equal $-1 \times -1 = \boxed{1}$ .

Note that if $a \in \{-1, 0, 1\}$ , then the set of four lines will include a vertical line with an undefined slope. So these three values cannot appear in set $A$ .

From this perspective, an interesting follow-up would be to repeat the problem with, say, $f(a) = \frac{\frac 1{\sqrt 3} + a}{1 - \frac 1{\sqrt 3}a} = \frac{1+\sqrt 3 a}{\sqrt 3 - a}$ based on $\tan 30^\circ = \frac 1{\sqrt 3}$ .