Five points in the plane are given, no three of which are collinear. Show that some four of them form a convex quadrilateral.
This question seems very simple but also at the same time very interesting. I have tried drawing many sets of $5$ dots and they all seem to have at least $1$ pair of $4$ form a convex quadrilateral, but how do I show that they all can't be concave? A concave quadrilateral is a quadrilateral in which at least one of its diagonals is not contained or is partly not contained in the quadrilateral.
Also, I am a bit confused on the case where $4$ points don't form a convex hull. We then have a triangle with $2$ points in it. How does this guarantee a convex quadrilateral?
Best Answer
If the convex hull $C$ of the five given points is a convex pentagon or a convex quadrilateral we are done. If $C$ is a triangle $C:=\triangle(A_0A_1A_2)$ then two of the five given points, say $P$ and $Q$, are lying in the interior of $C$. The line $\ell:=P\vee Q$ intersects two sides of $C$ in interior points. If $A_1$ and $A_2$ are two vertices of $C$ lying on the same side of $\ell$ then $A_1$, $A_2$, $P$, $Q$ are the vertices of a convex quadrilateral.
For the proof you may assume that $A_0=(0,0)$, $A_1=(1,0)$, $A_2=(0,1)$, and that $\ell$ intersects $[A_0A_1]$ at $(p,0)$ and $[A_0A_2]$ at $(0,q)$ with $p$, $q\in\ ]0,1[\ $. Choose any two points $P$, $Q\in\ell$ in the interior of $\triangle(A_0A_1A_2)$. It is then obvious by inspection that the four points $A_1$, $A_2$, $P$, $Q$ form a convex quadrilateral.