TITLE: Smallest number of points on plane that guarantees existence of a small angle
QUESTION [9 upvotes]: What is the smallest number $n$, that in any arrangement of $n$ points on the plane, there are three of them making an angle of at most $18^\circ$?
It is clear that $n>9$, since the vertices of a regular 9-gon is a counterexample. One can prove using pigeonhole principle that $n\le 11$. Take a edge of the convex hull of points. All the points lie to one side of this line. cut the half-plane into 10 slices of $18^\circ$ each. There can't be any points in the first and last slice. Thus, by pigeonhole principle some slice contains more than one point. So we have an angle of size at most $18^\circ$.
Now, for $n=10$, I can not come up with a counterexample nor a proof of correctness. Any ideas?

REPLY [4 votes]: Suppose there are $10$ points. Let us construct the convex hull of these points. It has at most $10$ sides. This means it must have an angle that is at most $180° - 360°/10 = 144°$. Let us choose the vertex corresponding to such angle as point $A$, and its 2 neighboring points along the convex hull as points $B$ and $C$. In other words, we have situation pictured here:

Now, from $7$ points inside the angle $\angle CAB$ let us draw 7 lines to point $A$. This will divide the angle into 8 smaller angles, and one of them will be at most $144°/8 = 18°$.
Therefore, since there is a counterexample for $n=9$ (a regular 9-gon), the answer is
$$n=10$$

$$GENERALIZATION$$
The smallest number $n$, that in any arrangement of $n$ points on the plane, there are three of them making an angle of at most $180°/k$ is $$n=k$$.
The proof is virtually the same as for the case $k=10$.