51 points lie inside a square of side 1. Prove that it's possible to draw a circle of radius $\frac17$ so that it covers at least 3 of the points.

Question

51 points lie inside a square of side 1. Prove that it's possible to draw a circle of radius $\frac{1}{7}$ so that it covers at least 3 of the points.

Here, I think that the pigeonhole principle may help, but I don't know what are the suitable holes.

Answer 1

The square of side length $1$ can be divided into $25$ squares of side length $\frac{1}{5}$, and by the pigeonhole principle one of these squares must contain three points.

Next, $\frac{2}{7}>\frac{\sqrt{2}}{5}$ since $\frac{4}{49}>\frac{4}{50}$, and so the square containing the three points can be fit inside a circle of radius $\frac{1}{7}$.

Answer 2

Hint: observe that a circle of radius $1/7$ contains an inscribed square with side $$ L = \frac{\sqrt{2}}{7} > \frac{1}{5}. $$ Hence you can cover the unit square with 25 of these squares, and so with 25 circles of radius $1/7$.