You’re given a deceptively simple geometry-meets-pigeonhole puzzle:

There are 51 ants scattered anywhere on a unit square.
You’re given a glass disk of radius $\frac{1}{7}$ .
Prove: No matter how the ants are placed, you can always position the glass to cover at least 3 ants.

Step 1: Divide and Conquer

The key is to break the unit square into manageable smaller regions.

Divide the square into a 5×5 grid of equal squares:

Each small square has side length $\frac{1}{5} = 0.2$
There are 25 sub-squares total.

This spatial partition gives us the perfect framework to apply the pigeonhole principle.

Step 2: Apply the Pigeonhole Principle

We have 51 ants placed anywhere in the square.

With only 25 sub-squares, some squares must contain multiple ants.

In fact:

If each of the 25 squares held at most 2 ants, that would total only $25 × 2 = 50$ ants.
But we have 51 ants → at least one square must contain 3 or more ants.

So, at least one of the 25 sub-squares contains at least 3 ants.

Step 3: Fit the Glass

Now we must show that a circle of radius $\frac{1}{7}$ is large enough to cover any three ants within such a small square.

Key Fact:

The diagonal of each small square is: $\sqrt{(0.2)^2 + (0.2)^2} = \sqrt{0.08} ≈ 0.2828$

So, the maximum distance between any two ants inside the same small square is less than 0.283.

A circle of radius $\frac{1}{7} ≈ 0.143$ has a diameter of $\frac{2}{7} ≈ 0.2857$ .

Thus, a circle of diameter 2/7 can cover the entire sub-square.

So: any three ants within that square can all fit under the glass.

Final Answer

Divide the square into 25 equal parts.
By the pigeonhole principle, one part must contain at least 3 ants.
That part fits inside a circle of radius $\frac{1}{7}$ , so the glass can cover them.

Reference

[1] Pigeonhole principle in geometric task