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

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?

• Is it an Euclidian plane or there is one with non zero curvature? – dmtri Feb 13 '19 at 7:22

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?
• Yes, I got it. We divide the region on the right of P into 10 distinct regions, we draw rays emerging from P to go to the right such that any two consecutive rays have an angle of $d = 18^o$ between them, 10*d would cover 180 degrees and hence would divide the region into 10 distinct regions, then we apply pigeonhole principle with 11 points as pigeons and these 10 regions as pigeonholes. Thanks! – Alan Feb 13 '19 at 7:03