Show that no matter how $12$ points are put on a plane, there are $3$ among them forming an angle not greater than $18^o$.

Question

Problem : Show that no matter how $12$ points are put on a plane, there are $3$ among them forming an angle not greater than $18^o$.

I am not getting any ideas in solving this problem. So, there will be $\binom{12}{3}= 220 $ triangles which means there will be a total of $660$ angles. We need to show that at least one of out of these $660$ angles will be less than or equal to $18^o$ degrees. How should we proceed now?

Answer 1

Hint: Consider a point $P$ having minimal $x$-coordinate among the $12$ provided points. What can you say about the angles from $P$ to the other $11$ points?