Putnam 2004, Problem B-4

Let $n$ be a positive integer, $n \geq 2$, and put $\theta = 2\pi/n$. Define points $P_k = (k, 0)$ in the $xy$-plane, for $k = 1, 2, \ldots, n$. Let $R_k$ be the map that rotates the plane counterclockwise by the angle $\theta$ about the point $P_k$. Let $R$ denote the map obtained by applying, in order, $R_1$, then $R_2, \ldots$, then $R_n$. For an arbitrary point $(x,y)$, find, and simplify, the coordinates of $R(x,y)$.

Solution:
**First solution:** Identify the $xy$-plane with the complex plane $\mathbb{C}$, so that $P_k$ is the real number $k$. If $z$ is sent to $z'$ by a counterclockwise rotation by $\theta$ about $P_k$, then $z' - k = e^{i\theta}(z - k)$; hence the rotation $R_k$ sends $z$ to $\zeta z + k(1 - \zeta)$, where $\zeta = e^{2\pi i/n}$. It follows that $R_1$ followed by $R_2$ sends $z$ to $\zeta(\zeta z + (1 - \zeta)) + 2(1 - \zeta) = \zeta^2 z + (1 - \zeta)(\zeta + 2)$, and so forth; an easy induction shows that $R$ sends $z$ to
$$\zeta^n z + (1 - \zeta)(\zeta^{n-1} + 2\zeta^{n-2} + \cdots + (n-1)\zeta + n).$$

Expanding the product $(1 - \zeta)(\zeta^{n-1} + 2\zeta^{n-2} + \cdots + (n-1)\zeta + n)$ yields $-\zeta^n - \zeta^{n-1} - \cdots - \zeta + n = n$. Thus $R$ sends $z$ to $z + n$; in cartesian coordinates, $R(x,y) = (x + n, y)$.

**Second solution:** (by Andy Lutomirski, via Ravi Vakil) Imagine a regular $n$-gon of side length 1 placed with its top edge on the $x$-axis and the left endpoint of that edge at the origin. Then the rotations correspond to rolling this $n$-gon along the $x$-axis; after the $n$ rotations, it clearly ends up in its original rotation and translated $n$ units to the right. Hence the whole plane must do so as well.

**Third solution:** (attribution unknown) Viewing each $R_k$ as a function of a complex number $z$ as in the first solution, the function $R_n \circ R_{n-1} \circ \cdots \circ R_1(z)$ is linear in $z$ with slope $\zeta^n = 1$. It thus equals $z + T$ for some $T \in \mathbb{C}$. Since $f_1(1) = 1$, we can write $1 + T = R_n \circ \cdots \circ R_2(1)$. However, we also have
$$R_n \circ \cdots \circ R_2(1) = R_{n-1} \circ R_1(0) + 1$$
by the symmetry in how the $R_i$ are defined. Hence
$$R_n(1 + T) = R_n \circ R_1(0) + R_n(1) = T + R_n(1);$$
that is, $R_n(T) = T$. Hence $T = n$, as desired.