Given 5 arbitrary points in the plane from which no 3 are collinear, prove that you can always choose 3 that are vertices of obtuse triangle.

Question

I have no idea how to prove it, any advice?

Answer 1

I assume that, after connection, the 5 points form a convex pentagon.

The sum of the 5 interior angles of that pentagon is … $540^0$.

Let any two of them be acute and the sum of them is then $180^0 – d$; where d is positive.

The sum of the remaining three will then be $(540 – (180 – d))^0$. That is, $(270 + d)^0$. This sum will be shared by the three. Clearly at least one of them will be obtuse.

Answer 2

The convex hull $C$ of the five points has $3$, $4$, or $5$ vertices.

If $C$ is a pentagon at least one of the angles of this pentagon has to be $\geq108^\circ$.

If $C$ has $\leq 4$ vertices there is at least one given point $p$ in the interior of $C$. By Caratheodory's theorem this point is in the convex hull of three of the given points. Connecting $p$ with these three points you will obtain at least one triangle having an angle $\geq120^\circ$.