Discrete Math: Placing Points in A Random Way

Question

I have a difficulty on how to approach this problem:

$51$ points are placed, in a random way, into a square of side $1$ unit. Can we prove that $3$ of these points can be covered by a circle of radius $\frac{1}{ 7}$ units ?

I thought of using Pigeonhole Principle, but I have no idea how to start doing this problem. Any help will be appreciated! Thanks :)

Answer 1

Let us consider that the square we are working on is $[0,1) \times [0,1)$ (see remark below).

Consider it as divided into $5 \times 5$ little squares. By the pigeonhole principle, at least one of these squares has 3 points (otherwise, if all the squares had at most 2 points in them, we would have $25 \times 2= 50$ points at most).

By Pythagoras theorem, the circumscribed circle to any small square has radius $\frac{\sqrt{2}}{10}=0.1414...$ which is (slightly) smaller than $\frac{1}{7}=0.1428...$.

Remark: strictly speaking, in order to have a partition of the big square, the small squares should be taken as $[\frac{k}{5},\frac{k+1}{5}) \times [\frac{\ell}{5},\frac{\ell+1}{5})$ (for $k,\ell = 0 \cdots 4$).