Show that if five points are picked on or in the interior of a square of side length 2, then there are at least two of these points no farther than √2 apart.

To solve this problem, we will use the Pigeonhole Principle and the fact that the diagonal of a square of side length 1 is √2.

Consider the square of side length 2, as shown below:

___________________
|         |         |
|    1    |    2    |
|_________|_________|
|         |         |
|    3    |    4    |
|_________|_________|

Divide the square into four smaller squares of side length 1, as shown above. By the Pigeonhole Principle, if we pick 5 points in the square, at least two of them must be in the same smaller square.

Now consider the diagonal lines that connect opposite corners of each smaller square, as shown below:

___________________
|         |         |
|    1    | /   2   |
|_________|_________|
|  /   3  |/   4    |
|_________|_________|

Each diagonal has length √2, which is less than 2. Therefore, if two points are in the same smaller square, they are no more than √2 apart. And since there are only four smaller squares, if we pick 5 points in the square, there must be at least two points in the same smaller square, which are no more than √2 apart.

Therefore, we have shown that if five points are picked on or in the interior of a square of side length 2, then there are at least two of these points no farther than √2 apart.