# Prove existence of a triangle with least angle $\leq 30$ degrees

Problem: Let $A$ be a set of $6$ points in a plane such that no $3$ are collinear. Show that there exist 3 points in $A$ which form a triangle having an interior angle not $30$ degrees.

I am supposed to use the Pigeonhole Principle for this problem but I am unable to see how to do that here. I considered the triangle with vertices in the point set and least area and observed that no other point in $A$ can be inside this triangle. But I couldn't make use of it.

Also, since the points don't necessarily form a convex set I am unable to use the angle restrictions that we have on convex polygon.

I have a hunch that we need to show that some mean is less than 30 (maybe we can show that the sum of smallest angles taken over all the 20 triangle s is less than 600?)

If all six points lie on the convex hull, they form a hexagon with interior angle sum $(6-2)\pi=4\pi$, so at least one interior angle is at least $\frac23\pi$, so at least one of the other two angles in the corresponding triangle is at most $\frac\pi6$.
Else, one point lies in a triangle formed by three others. At least one angle of that triangle is at most $\frac\pi3$, and the line to the interior point divides it into two parts, at least one of which is at most $\frac\pi6$.
• Alternatively for the 1st para: the interior angle of at least one vertex $P$ is at most $2\pi/3$. The 3 diagonals from $P$ divide that angle into 4 angles, at least one of which is at most $(2\pi/3)/4=\pi/6$. – Rosie F Jun 6 '16 at 9:51