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.
