Consider a polygonal line $P_0P_1\ldots P_n$ such that $\angle P_0P_1P_2=\angle P_1P_2P_3=\cdots=\angle P_{n-2}P_{n-1}P_{n}$, all measured clockwise. If $P_0P_1>P_1P_2>\cdots>P_{n-1}P_{n}$, $P_0$ and $P_n$ cannot coincide.
While the result is remarkable in itself, the following proof that appears in Titu's 'Mathematical Olympiad Challenges' is just mind blowing :
Let us consider complex coordinates with origin at $P_0$ and let the line $P_0P_1$ be the x-axis. Let $\alpha$ be the angle between any two consecutive segments and let $a_1>a_2>\cdots>a_n$ be the lengths of the segments. If we set $z=e^{i(\pi-\alpha)}$, then the coordinate of $P_n$ is $a_1+a_2z+\cdots+a_n z^{n-1}$. We must prove that this number is not equal to zero.
Using the Abel Summation lemma, we obtain, $$a_1+a_2z+\cdots+a_{n-1}z^n=(a_1-a_2) + (a_2-a_3)(1+z) + \cdots + a_n(1+z+\cdots+z^{n-1})$$
If $\alpha$ is zero, then this quantity is a strictly positive real number, and we are done.
If not, multiply by $(1-z)$ to get $(a_1-a_2)(1-z) + (a_2-a_3)(1-z^2) + \cdots + a_n(1-z^n)$. This expression cannot be zero. Indeed, since $\mid z\mid=1$, by the triangle inequality, we have, $$\mid (a_1-a_2)z + (a_2-a_3)z^2 + \cdots + a_n z^n \mid < (a_1-a_2) + (a_2-a_3) + (a_3-a_4) + \cdots + a_n$$
The conclusion follows.
I was wondering if one can prove this synthetically, without using complex numbers. Any alternate proof or suggestion in the direction of an alternate proof is welcome.
If the polygon was to return to the origin, then $n\alpha=2\pi$, because the returning line must make an angle of $\alpha$ with the starting line.
Place each edge of the polygon at the origin at an angle of $k\alpha$, starting on the positive $x$-axis, and working clockwise. Each line segment below the $x$-axis has a counterpart line segment above the $x$-axis, and the $y$-value is strictly larger, therefore the polygon cannot return to the origin.