# Problem 1325 - A surprising polygon formula

## Problem of the Week from Stan Wagon.

By Vamshi Jandhyala in mathematics

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 ðŸ˜Š. A simple trigonometric proof is given below. It is fairly straightforward ðŸ˜‰ 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.