IMC 2011 · Day 2 · P10

The operations $f_i$ are rational maps on the $2(n-1)$-dimensional phase space of coordinates of the vertices $A_1, \dots, A_{n-1}$. To show that $(f_1 \circ f_2 \circ \dots \circ f_{n-1})^n$ is the identity, it is sufficient to verify this on some open set. For example, we can choose a neighborhood of the regular polygon, then all intermediate polygons in the proof will be convex.

Consider the operations $f_i$. Notice that (i) $f_i \circ f_i = id$ and (ii) $f_i \circ f_j = f_j \circ f_i$ for $|i - j| \ge 2$. We also show that (iii) $(f_i \circ f_{i+1})^3 = id$ for $1 \le i \le n-1$.

The operations $f_i$ and $f_{i+1}$ change the order of side lengths by interchanging two consecutive sides; after performing $(f_i \circ f_{i+1})^3$, the side lengths are in the original order. Moreover, the sums of opposite angles in the convex quadrilateral $A_{i-1} A_i A_{i+1} A_{i+2}$ are preserved in all operations. These quantities uniquely determine the quadrilateral, because with fixed sides, both angles $\angle A_1 A_2 A_3$ and $\angle A_1 A_4 A_3$ decrease when $A_1 A_3$ increases. Hence, property (iii) is proved.

In the symmetric group $S_n$, the transpositions $\sigma_i = (i, i+1)$, which from a generator system, satisfy the same properties (i–iii). It is well-known that $S_n$ is the maximal group with $n-1$ generators, satisfying (i–iii). In $S_n$ we have $(\sigma_1 \circ \sigma_2 \circ \dots \circ \sigma_{n-1})^n = (1, 2, 3, \dots, n)^n = id$, so this implies $(f_1 \circ f_2 \circ \dots \circ f_{n-1})^n = id$.