Sum of distance to odd vertices = Sum of distance to even vertices in regular polygon

Question

Let $P_1P_2\dots P_{2m+1}$ be a regular polygon, and let $K$ be a point on the minor arc $P_1P_{2m+1}$ of its circumcircle. Show that $$\sum_{i~\text{odd}}KP_i=\sum_{i~\text{even}}KP_i.$$

I have tried using complex numbers on this problem (setting $P_k$ as the roots of $x^{2m+1}-1=0$, but to no avail.

Perhaps there is a Euclidean approach?

Answer 1

Let $O$ be the center of the circumscribed circle and $r$ its radius. Let $\beta = \frac{1}{2}\widehat{P_1 O P_2}=\frac{\pi}{2m+1}$ and $\theta = \frac{1}{2}\widehat{KOP_1}$ with $0 \le \theta \le \beta\,$.

A chord that subtends an angle $\varphi$ has a length of $2 r \sin \frac{\varphi}{2}\,$, so $KP_k = 2 r \sin\big(\theta + (k-1)\beta\big)$ for $1 \le k \le 2m+1$. Then:

$$ S_{odd} = \sum_{k=0}^{m} KP_{2k+1} = 2r\,\sum_{k=0}^{m} \sin\big(\theta + 2k\beta\big) $$

$$ S_{even} = \sum_{k=1}^{m} KP_{2k} = 2r\,\sum_{k=1}^{m} \sin\big(\theta + (2k-1)\beta\big) $$

Using the following identity for sums of sines with arguments in an arithmetic progression:

$$ \sin \varphi + \sin(\varphi+\alpha) + \cdots + \sin(\varphi+ n \alpha) = \cfrac{\sin\frac{(n+1)\alpha}{2} \sin\big(\varphi+\frac{n\alpha}{2}\big)}{\sin \frac{\alpha}{2}} $$

• with $\varphi = \theta, \alpha=2\beta, n=m\,$: $\;\;S_{odd} = 2r\,\cfrac{\sin\big((m+1)\beta\big)\,\sin\big(\theta+m \beta\big)}{\sin \beta}$

• with $\varphi = \theta+\beta, \alpha=2\beta, n=m-1\,$: $\;\;S_{even} = 2r\,\cfrac{\sin\big(m \beta\big)\,\sin\big(\theta+m \beta\big)}{\sin \beta}$

But $\sin((m+1)\beta)=\sin(m \beta)$ since $(m+1)\beta + m\beta = (2m+1)\beta=\pi$ therefore $S_{odd}=S_{even}$.

As a comment, the case $m=1$ reduces to showing that $KP_2=KP_1+KP_3$ for an equilateral triangle $P_1P_2P_3$ with $K$ on the arc $P_1P_3$. This is a well known result, which follows directly from Ptolemy's theorem for the cyclic quadrilateral $KP_1P_2P_3$. It is tempting to think that maybe McDougall's Generalization of Ptolemy's Theorem could lead to a more geometric proof for OP's question here, though that's not immediately obvious.

Answer 2

I've figure out a particularly nice solution using nothing other than standard Ptolemy's theorem.

Let $a=P_1P_2$ and $b=P_1P_3$. Apply Ptolemy's theorem to quadrilaterals of the form $KP_{i-1}P_iP_{i+1}$, where $i$ ranges over the integers $\pmod{2m+1}$. We get the equalities $$\begin{align*}a\cdot KP_2&=b\cdot KP_1+a\cdot KP_{2m+1}\\ b\cdot KP_2&=a\cdot KP_3+a\cdot KP_1\\ a\cdot KP_2+a\cdot KP_4&=b\cdot KP_3\\ b\cdot KP_4&=a\cdot KP_3+a\cdot KP_5\\&~~\vdots\\ a\cdot KP_{2m}&=a\cdot KP_1+b\cdot KP_{2m+1}\end{align*}$$ Adding these gives $$(2a+b)\sum_{i~\text{even}}KP_i=(2a+b)\sum_{i~\text{odd}}KP_i.$$