Armo wrote: In an $n$-gon $A_{1}A_{2}\ldots A_{n}$, all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation \[a_{1}\geq a_{2}\geq \dots \geq a_{n}. \] Prove that $a_{1}=a_{2}= \ldots= a_{n}$. Overlay the complex plane with origin $A_{1}$, rotating and scaling so that $A_{2}= 1+0i$. WLOG, the entire polygon lies on the complex half-plane with positive imaginary coefficient. Let $z_{i}$ denote the complex number $A_{i+1}-A_{i}$, where $A_{n+1}= A_{1}$ and the vertices refer to the corresponding complex number. It is clear that $z_{1}+z_{2}+\cdots+z_{n}= 0$. Now, as $im(z_{1}) = 0$, consider $im(z_{n}+z_{2})+im(z_{n-1}+z_{3})+\cdots+im(z_{2}+z_{n}) = 0$. By the side-order length, we must have $|z_{i}| \le |z_{j}|$ for all $1 \le i < j \le n$. But $z_{n-k}$ and $z_{k+2}$ make equal and opposite angles with the real axis. It follows that $im(z_{n-k}+z_{k+2}) \ge 0$ with equality iff $|z_{n-k}| = |z_{k+2}|$. As the sum is 0 when added over all $k$, it must be that each term is 0, hence, the polygon is regular.

Suppose $A_1A_2=A_kA_{k+1}$, then construct a regular n-gon $A_1A_2...A_{k+1}A'_{k+2}...A'_n$. Denote $A_{j}A_{j+1}-A_{j+1}A_{j+2}$ by $x_j$, then $d(A_{j+1}A_{j+2},A'_{j+1}A'_{j+2})=\sum_{i=k+1}^j x_i\sin\frac{2\pi}{n}$. Clearly contradiction occurs.

This IMO 1963,problem 3. The original solution is purely geometrical. Here's another approach. I'll use the following result: Let $a_{1},a_{2},\ldots a_{n}$ be positive real numbers and let $\varepsilon$ be a primitive $n$th root of the unity, e.g. $\varepsilon=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}.$ If the sides of an equiangular polygon have lengths $a_{1},a_{2},\ldots,a_{n}$ (in counterclockwise order) then \[ a_{1}+a_{2}\varepsilon+a_{3}\varepsilon^{2}+\ldots+a_{n}\varepsilon^{n-1}=0. \] The solution of the problem is obtained from the following: Lemma. Let $P(X)=a_{1}+a_{2}X+\ldots+a_{n}X^{n-1},$ where $a_{1}\geq a_{2}\geq\ldots\geq a_{n}>0.$ If $\alpha$ is a root of $P,$ then $|\alpha |\geq1,$ and $|\alpha|=1$ only if $a_{1}=a_{2}=\ldots=a_{n}. $ Proof. We have \[ a_{1}+a_{2}\alpha+\ldots+a_{n}\alpha^{n-1}=0. \] If we multiply this equality with $\alpha-1,$ we obtain \[ -a_{1}+\alpha(a_{1}-a_{2})+\alpha^{2}(a_{2}-a_{3})+\ldots+\alpha^{n-1}% (a_{n-1}-a_{n})+a_{n}\alpha^{n}=0, \] or, equivalently, \[ a_{1}=\alpha(a_{1}-a_{2})+\alpha^{2}(a_{2}-a_{3})+\ldots+\alpha^{n-1}% (a_{n-1}-a_{n})+a_{n}\alpha^{n}. \] Now, suppose that $|\alpha|\leq1.$ We obtain \[ a_{1}=|\alpha(a_{1}-a_{2})+\alpha^{2}(a_{2}-a_{3})+\ldots+\alpha^{n-1} (a_{n-1}-a_{n})+a_{n}\alpha^{n}|\leq \] \[ \leq|\alpha|(a_{1}-a_{2})+|\alpha|^{2}(a_{2}-a_{3})+\ldots+|\alpha |^{n-1}(a_{n-1}-a_{n})+a_{n}|\alpha|^{n}\leq \] \[ \leq(a_{1}-a_{2})+(a_{2}-a_{3})+\ldots+(a_{n-1}-a_{n})+a_{n}=a_{1}. \] Consequently, all inequalities must be equalities. Because $\alpha \notin\mathbb{R},$ this is possible only if $a_{1}=a_{2}=\ldots=a_{n};$ that is, if the polygon is regular.

Well, this seems to be an easy problem, however, I may be incorrect with my method of proof. We know that in an $n-gon$, the sum of all the interior angles is $(n-2)180$. Draw a diagonal from each vertex to it's opposite vertex, creating $n$ iscosolese triangles. Let the tip of all triangles touching the centre have an angle $\alpha$. It is now obvious that $\alpha=180-\frac{(n-2)180}{n}=360/n$. Since isoscelese triangles are all equal to one another, all sides are equal. Masoud Zargar

And what do you understand by opposite vertex when the polygon has an even number of sides?

Valentin Vornicu wrote: And what do you understand by opposite vertex when the polygon has an even number of sides? Do you mean: What do you understand when they have an odd number of sides? If it has even sides, we will be able to create the triangles. However, if we don't...I need to think abotu this Masoud Zargar

Let $A_1A_2\dots A_n$ be the vertices of the n-gon, and $A_1A_2=a_1$, $A_2A_3=a_2$,..., $A_{n-1} A_n=a_{n-1}$ , $A_nA_1=a_n$. Now project the vertices into the bisector of the angle $\angle A_{n-1}A_nA_1$. It's evident, given the equality of inner angles of the polygon and the inequalities for the lenghts of the sides, that the projection of $A_1A_2\dots A_{\lfloor\frac{n+2}2\rfloor}$ is strictly longer than the projection of the polygonal $A_1A_nA_{n-1}\dots A_{\lfloor\frac{n+3}2\rfloor}$ unless $a_1=a_2=\dots=a_n$. QED