From 5 random points on a plane, every 3 of them aren't on the same line.
Prove that we can always pick 3 of them which are the vertices of a obtused triangle.
I assume it's a variation of the Pigeonhole principle, but I'm have no idea how to do it.
2026-04-02 02:29:18.1775096958
Random 5 points, we can choose 3 which are creating a obtused triangle.
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Consider the convex hull of the five points. If that hull consists of all five points, it makes up a pentagon, which means that the sum of its interior angles is $(5 - 2)\cdot 180^\circ = 540^\circ$. Therefore, at least one of the interior angles has to be an obtuse one, which gives us an obtuse triangle. If The hull consits of four points or less, you have at least one point inside it. Connect that point to each of the hull's points. At least one of the angles with the inner point of the vertext has to be obtuse, unless the hull consists of four points and each of the angles is a right angle, which cannot be the case, since then the inner point would lie on the diagonals of the hull, which would mean that 3 points are colinear, which cannot be the case by our assumption.