Assume that for any pair of vertices $P_i$ and $P_j$ , there exists a vertex $P_k$ of the polygon such that $∠P_i P_k P_j = \pi/3.$ 
Let $P_1 P_2 \dots P_n$ be a convex polygon in the plane. Assume that for any pair of vertices $P_i$ and $P_j$ , there exists a vertex $P_k$ of the polygon such that $∠P_i P_k P_j = \pi/3.$
  Show that $n = 3$

Taken from Romanian Mathematical Olympiad, 2000
Alright, how can you show this one? There may be many solutions of it
 A: There is a basic observation about the relationship between the angles of a triangle and the lengths of its sides: 
$\textbf{Fact 1.}$ If $\triangle ABC$ is with $\angle C \geq \pi/3$, then it's either equilateral, or has a side shorter than $AB$. Similarly if $\angle C\leq \pi/3$ then it's either equilateral or has a side longer than $AB$.
You can deduce this using the law of cosines, writing $c^2 = a^2 - 2ab\cos\angle C + b^2$. Alternatively, you can observe that if $\angle C\geq \pi/3$ then either $\angle A\leq \pi/3$ or $\angle B\leq \pi/3$, so there the smallest angle is at most $\pi/3$, and since smaller angles have smaller sides opposite them we get the conclusion (analogously for the $\angle C\leq \pi/3$ case).
Now let $d$ be the shortest distance between any two points from the polygon, realized by the segment $P_iP_j$. Then choosing $P_k$ with $\angle P_iP_kP_j=\pi/3$ shows that $P_iP_jP_k$ is equilateral. Similarly if $Q_iQ_j$ is the longest distance $D$ we get an equilateral triangle $Q_iQ_jQ_k$.
Since $P$ is convex, the points that are not $P_i,P_j,P_k$ must lie in union of the three infinite regions determined by taking one of the angles of $\triangle P_iP_jP_k$ and removing $\triangle P_iP_jP_k$ from it.
Now, if say $Q_i,Q_j$ are in the same region, say corresponding to the angle at $P_i$, then $\angle Q_iP_iQ_j<\pi/3$ and thus one of $P_iQ_i,P_iQ_j$ is longer than $Q_iQ_j$, contradiction.
Thus, each of $Q_i,Q_j,Q_k$ is in its own region (and they can possibly coincide with vertices of $P_iP_jP_k$). WLOG $Q_i$ is in the region determined by $\angle P_i$. Then $\angle P_jQ_iP_k\leq\pi/3$ for otherwise we get a distance shorter than $d$. But since $P$ is convex, $Q_j,Q_k$ must be inside $\angle P_jQ_iP_k\leq \pi/3$, and since $\angle Q_jQ_iQ_k=\pi/3$, they must in fact lie on the rays $\vec{Q_iP_j}$ and $\vec{Q_iP_k}$. Since $P$ is convex, there are no three vertices on the same line, so the only way that can happen is when $\{P_j,P_k\}=\{Q_j,Q_k\}$. 
Thus $D=d$, and hence the distances between all pairs of vertices of $P$ are the same, which can only happen when $n=3$.
A: 
I like pictures! This will have less rigor, but drawing this gave me nice intuition for why the theorem holds. And, reading amakelov's answer, the logic is also slightly different.
Let points A and B be a pair of vertices in the polygon with maximal distance between them. Let C be a vertex that forms the appropriate $\pi/3$ triangle. If either of the legs is shorter than the other, then the longer leg is longer than AB, a contradiction (Fig 1); so ABC must be equilateral (Fig 2).
Let's attempt to add another vertex. Without loss of generality, we can put it farthest from C. But this new vertex E can't be farther from C than A and B are from C; it also must stay convex. So E is confined to the ⌓ between the green triangle and lime green circle. Further, without loss of generality, lets put E closest to A. And in fact, we'll require it to be the closest vertex of the entire polygon to A -- so any subsequent vertices we add have to be farther from A than E.
But E must have a $\pi/3$ completion with A. Convexity constraints from A and B are drawn in lime green straight lines: the new vertex must be in the narrow regions. The new vertex must also form $\pi/3$, placing it on one of the blue circles. Finally, it must be further from A than E, which excludes the portion of the blue circles on the left of the orange line.
The straight lime green lines will always be nearly parallel to AB, while the blue circles passing through E go through at much more vertical angles, so there is no region in the AB circular segment ⌓ that could host a new vertex -- looking on the far side of the circles, it is blocked off by the orange line. The only other possibility is that the new vertex is in the CA region. However, the orange line excludes everything in the right half of the blue circles, so clearly the allowed part of the blue arc will never be inside the CA ⌓. Thus AC cannot have a valid 5th vertex.
Wait, why I didn't have a vertex named D? Hmm.
