# Five Points in a Square

There are 5 points in a square 1×1. Show that 2 of the points are within distance 0.75.

**SOLUTION**

Split the unit square into 4 small squares with side lengths 0.5. At least one of these squares will contain 2 of the points. Since the diagonals of the small squares have lengths less than 0.75, these 2 points must be within such distance.

**email**.

Anyway, it might be interesting to add the constraint that the points have to be on the sides or corners of the unit square, in which case the farthest apart I could find was about 0.647111. The challenge could be to derive the exact solution.

That’s a good problem, I will think about it! Thanks:)

The farthest apart I can find is about 0.647111. I can’t mathematically prove that it’s the farthest though.

Hi Mel. The optimal solution is to put four points in the vertices of the square and one in the center. The proof that you can’t do better is in the solution underneath the problem:)

Yes, I see that your solution is half of the square root of two and therefore better than what I showed, and it probably is the optimum solution, but I don’t see any proof of that.

The proof that you can’t do better than sqrt(2) is that there are always 2 points within a 1×1 square. Therefore, the distance between them is at most the length of the diagonal which is sqrt(2).