Property of Vectors of an n-gon Let ${P_1,..., P_n}$ be the vertices of a regular n-gon in the plane, and $O$ it's center; show without computation or coordinates that ${\overrightarrow {OP_1} + \overrightarrow {OP_2} + ... + \overrightarrow {OP_n}} = 0$
a) if n is even
b) if n is odd
My attempt at solving this problem makes use of the fact that if we have any closed polygon in the plane which does not cross itself with vertices ${P_1,..., P_n}$, the following holds true: ${\overrightarrow {P_1P_2} + \overrightarrow {P_2P_3} + ... + \overrightarrow {P_nP_1}} = 0$.
Say we constructed vectors ${\overrightarrow {A_1}, \overrightarrow {A_2},... \overrightarrow {A_n}}$ where $\overrightarrow {A_1}$ is a vector with the same magnitude as $\overrightarrow {P_1P_2}$ rotated $\theta°$ counter-clockwise to it in the plane, $\overrightarrow {A_2}$ is a similar vector for $\overrightarrow {P_2P_3}$ etc, then we can say that ${\overrightarrow {A_1} + \overrightarrow {A_2} + ... + \overrightarrow {A_n}} = 0$.
If we arbitrarily scaled the magnitudes of all the vectors ${\overrightarrow {A_1}, \overrightarrow {A_2},... \overrightarrow {A_n}}$ then the fact above still holds. If we rotated these vectors all by the same angle in the same direction then the fact above still holds.
Using this logic, I think we can conclude that ${\overrightarrow {OP_1} + \overrightarrow {OP_2} + ... + \overrightarrow {OP_n}} = 0$ as the vectors formed by joining $O$ to ${P_1,..., P_n}$ all have the same magnitude and the heads of these vectors are touching the vectors ${\overrightarrow {P_1P_2},  \overrightarrow {P_2P_3}, ..., \overrightarrow {P_nP_1}} = 0$., ie they are scaled down by some factor and rotated by some angle $\theta$ which is dependent on the number of sides of the regular n-gon $n$.
I now do not know if this logic is sound and it doesn't explain why the question is asking for the case when $n$ is a) even and b) odd.
 A: This argument doesn't separates cases where the number of vertices is even or odd, but if $R$ denotes the rotation about $O$ by $\frac{1}{n}$ of a turn, then $R$ cyclically permutes the vertices, and therefore preserves the sum
$$
\sum_{k=1}^{n} OP_{k}.
$$
Since $O$ is the unique fixed point of $R$, the sum must be $O$.
A: $n$ is even is simple, add opposite vectors together.
For $n$ odd, assume w.l.o.g  that:
$$\sum\limits_{i=1}^n \overrightarrow {OP_i} \gt 0$$
We know that it's reflection produces a $2n$-gon, and that the sum of the reflected vectors is negative.
By the Intermediate Value Theorem, rotating the original vectors must produce a value of $0$ somewhere, however we have preserved the orignal $n$-gon, and this is a contradiction.
A: We can assume that $P_1 = \omega = e^{\frac{2\pi i}{n}}$. Then
$$\vec{OP_1} + \vec{OP_2} + \cdots +\vec{OP_{n}} = \omega + \omega^2 + \cdots + \omega^{n} $$
Noting that $\omega^n = 1$ and hence $1+\omega + \cdots + \omega^{n-1} = 0$, the result follows.
