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I was fascinated by this problem from the first moment I saw it.

Let $P$ be a convex polygon which has no two sides which are parallel. Each side $A_iA_{i+1}$ has a furthest away point $C_i$. Prove that $$ \sum_{i=1}^n \angle A_iC_iA_{i+1}=\pi$$

This was given in 2005 at a contest for junior level olympiad students in Romania.

I've just compiled and understood a solution, which is quite complicated (you can see it here)

Is there a simple solution for this problem using some kind of geometric transformation, or clever reduction to a particular case? (Obviously, the case of a cyclic polygon is very easy)

At first sight it didn't seem intuitive for me that for any polygon that sum of angles should be constant. Maybe some deeper insight into the problem or a heuristical argument could explain the result better...

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Reduce to: All points $C$ are distinct.

If two sides have the same point $C$, then there are two neighbouring sides having the same point $C$. Replace these two sides $A_kA_{k+1}$ and $A_{k+1}A_{k+2}$ which reduces the number of points by 1.

Reduce to: The number of vertices is odd ($2k+1$) and $C(A_iA_{i+1}) = A_{i+k+1}$ for all $i$

Regard a point $C(A_iA_{i+1})$ and two portions of the polygon boundary between $C$ and the side. The sides on one portion have to take all their (distinct!) $C$ points in the other portion, which implies that their number is the same and $C$ is the numerically opposite point of the polygon.

Turn the line $A_iA_{i+k}$ around once which gets total angle $\pi$.

Start with $A_0A_k$ to $A_0A_{k+1}$ around $A_0$, then to $A_1A_{k+1}$ around $A_{k+1}$ and so on.

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That is a very nice interpretation. Up to a point, that is the solution I found, also, but the last part is nicer. – Beni Bogosel Nov 29 '12 at 21:34
@BeniBogosel I think that there should be a solution using the polygon enveloped by the parallels of the sides through their $C$, but I couldn't make it work. – Phira Nov 29 '12 at 21:42

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