Problem: Given any set $S$ of $9$ points within a unit square, show that there always exist $3$ distinct points in $S$ such that the area of the triangle formed by these $3$ points is less than or equal to $\frac{1}{8}$
I am supposed to do this using the pigeonhole principle. I used the usual technique of partitioning the square to conclude that there is a triangle with area less than $\frac{1}{4}$ but I am unable to improve the bound.
Please help.
First divide the unit square into four squares with sides $s$ of length $\frac{1}{2}$.
Now given $9$ points and $4$ squares, by the pigeon hole principle at least one square $S$ which contains three of those points. Those three points form a triangle whose area $A$ is bounded by half the area of $S$ (see here for a proof). Since the sides $s$ of $S$ have length $\frac{1}{2}$, we have that $$ A\leq \frac{1}{2}s^2= \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$$
