Suppose we have area bound by $0\leq x\leq 2$,$0\leq y\leq 2$

We choose 17 points, there should be at least two with distance max $d\leq\frac{1}{\sqrt{2}}$

Tried to count area of $A_{17circles}=(\frac{d}{2})^2\pi*17=6.67588438887831$

Then to take square $A_{square}=(\frac{d}{2}+2+\frac{d}{2})^2=7.32842712474619$

There is also each corner to be rounded $A_{removecorners}=d^2-(\frac{d}{2})^2\pi=0.107300918301276$

Which results in $A_{square}-A_{removecorners}=7.22112620644491$

Still $7.22112620644491>6.67588438887831$ so I cant dismiss argument that one is able to fit 17 points so, that $max_{distance}\geq d$

How to dismiss this argument? How to prove that if we put 17 points into 2 by 2 box, some of them bound to be closer than $\frac{1}{\sqrt{2}}$

up vote 2 down vote accepted

Hint: Divide the $2\times 2$ square into $16$ equal little squares, by drawing three equally spaced horizontal lines, and three vertical lines. Then use the Pigeonhole Principle.

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