# Problem 1325 - A surprising polygon formula

Problem of the Week from Stan Wagon.

mathematics
Published

July 9, 2021

## Problem

For a regular polygon with $$n$$ sides ($$n$$ even) and side-length $$2$$, prove that the sum of the squares of the segments along a vertical line as in the diagram is $$n$$. Remarkably, the same identity works for an odd number of sides. But the geometry in the two cases is a little different, so the main problem is for the even case only.

## Solution

A regular polygon with $$n$$ sides where $$n$$ is even has $$n=4k$$ or $$n=4k+2$$ sides. The coloured line segments are the projections of the polygon sides onto the $$y$$ axis.

### Case: $$n=4k$$

Let a regular polygon with $$n=4k$$ sides and side length $$2$$ be oriented such that its center is at the origin and both the $$x$$ and $$y$$ axis are lines of symmetry.Let the sides of the polygon be numbered from $$1$$ to $$4k$$ in anti clockwise direction starting from the side with one vertex on the positive $$x$$ axis and another vertex in the first quadrant. Let $$x_i$$ and $$y_i$$ be the lengths of the projections of side $$i$$ on the $$x$$ and $$y$$ axis respectively. We have $$x_i^2+ y_i^2 = 2^2$$ for all $$i \in \{1, \dots, 4k\}$$. We also have $$x_{i+k} = y_i$$ and $$y_{i+k} = x_i$$ because the side $$i+k$$ is perpendicular to side $$i$$ (the angle between them is $$\frac{2\pi}{4k}k$$).

The sum we are interested in is

$\sum_{i=k+1}^{3k} y_i^2 = \sum_{i=1}^{2k} y_i^2$

due to symmetry.

We also have

\begin{aligned} \sum_{i=1}^{2k} y_i^2 &= \sum_{i=1}^{k} y_i^2 + \sum_{i=k+1}^{2k} y_i^2 \\\\ &= \sum_{i=1}^{k} y_i^2 + \sum_{i=1}^{k} x_i^2 \\\\ &= \sum_{i=1}^{k} x_i^2 + y_i^2 \\\\ & = k \cdot 2^2 = 4k = n \end{aligned}

This completes the proof of the formula for the case where the polygon has $$n=4k$$ sides.

### Case: $$n=4k+2$$

Haven’t been able to find an elegant geometric proof yet 😊. A simple trigonometric proof is given below. It is fairly straightforward 😉 to see that the required sum of squares of projections on the $$y$$-axis is

\begin{aligned} &2 \sum_{i=1}^{k} \left(2 \sin{\frac{(2i-1)\pi}{4k+2}} \right)^2 + 4 \\\\ &= 2\sum_{i=1}^k \left(2 - 2\cos{\frac{(2i-1)\pi}{2k+1}} \right) + 4 \\\\ & = 4k + 4 - 4\sum_{i=1}^{k} \cos{\frac{(2i-1)\pi}{2k+1}} \\\\ & = 4k + 4 - \frac{2}{\sin{\frac{\pi}{2k+1}}} \sum_{i=1}^{k} \sin{\frac{2i\pi}{2k+1}} - \sin{\frac{2(i-1)\pi}{2k+1}}\\\\ & = 4k + 4 - 2 \frac{\sin{\frac{2k\pi}{2k+1}}}{\sin{\frac{\pi}{2k+1}}} \\\\ & = 4k + 2 \end{aligned}

This completes the proof of the formula for the case where the polygon has $$n=4k+2$$ sides.