Can You Squeeze the Squares?

Fiddler

There’s a square board with side length A. Your friend cleverly places a unit square on the board and challenges you to place another unit square on the board—without moving the first one—so that it too is entirely on the board and the squares don’t overlap. (The unit squares can touch each other.)

Alas, it’s impossible for you to do so! But there’s some minimum value of A for which you can always place a second unit square on the board, no matter how cleverly your friend places the first one.

What is this minimum value of A?

Solution

The most strategic position for your friend is to place it in the center. Any offset from the center creates relatively more room in one or two corners.

The most strategic place for you to find a place to fit your square is in a corner.

Answer

If $A \lt 3$, there is no space. If $A = 3$, you will always be able to fit a square in a corner.

Extra Credit

Now there’s another square board with side length B. This time, your friend cleverly places three unit squares (which can touch but not overlap) and issues a similar challenge, asking you to place one more unit square on the board.

Once again, it’s impossible for you to do so! But there’s some minimum value of B for which you can always place a fourth unit square on the board, no matter how cleverly your friend places the first three squares.

What is this minimum value of B?

Solution

Consider the tightest circle containing any rotational possibility of your friend's three squares, centered at the center of the square board.

Note that if your friend's three squares move away from the center, one corner will always remain viable for you to place a square.

Even if centered, rotating all three squares together leaves one or two corners.

Answer

If $B \lt 2 + \sqrt{2}$, there is no space. If $B = 2 + \sqrt{2}$, you will always be able to fit a square in a corner.

Rohan Lewis

2025.08.04

Code can be found here.