Covering Three of 51 Ants With a Small Glass
Fifty-one ants are sitting on a square table whose side length is $1$ (all ants are inside or on the boundary of the unit square). You have a circular glass whose base is a disc of radius $1/7$.
Show that, no matter how the 51 ants are arranged, you can always place the glass so that it covers at least three ants (an ant on the boundary of the disc counts as covered).
Hints
- With 51 ants and a small glass, think pigeonhole: divide the square into regions so that some region is forced to contain at least three ants.
- A $5 \times 5$ grid of squares of side $1/5$ gives 25 regions for 51 ants, so some small square holds at least $\lceil 51/25 \rceil = 3$ ants.
- A square of side $1/5$ has diagonal $\sqrt{2}/5 \approx 0.283$, which is less than the glass diameter $2/7 \approx 0.286$. So the glass centred on that small square covers it entirely.
Worked Solution
How to Think About It: You cannot control where the ants are, so the argument must be an existence proof: partition the square into pieces small enough that the glass can cover any one of them, and use counting to show some piece has at least three ants. The only tension is choosing the grid so that both conditions hold with the given radius.
Quick Estimate: We want cells that fit inside a disc of diameter $2/7 \approx 0.2857$. A square of side $s$ fits if its diagonal $s\sqrt{2} \le 2/7$, i.e. $s \le 0.202$. Side $1/5 = 0.2$ works, giving 25 cells, and $51 > 2 \times 25$ forces a triple.
Formal Solution:
*Step 1 -- Partition.* Divide the unit square into a $5 \times 5$ grid of $25$ closed squares of side $1/5$. Every ant lies in at least one of these cells (an ant on a grid line belongs to more than one; assign it to any one of them).
*Step 2 -- Pigeonhole.* If every cell contained at most $2$ ants, there would be at most $2 \times 25 = 50$ ants. There are $51$, so some cell contains at least $3$ ants. (Formally, some cell holds at least $\lceil 51 / 25 \rceil = 3$ ants.)
*Step 3 -- The glass covers a cell.* A square of side $1/5$ has diagonal $\frac{\sqrt{2}}{5} \approx 0.2828$. Its circumscribed circle has radius half the diagonal, $\frac{\sqrt{2}}{10} \approx 0.1414$. The glass has radius $\frac{1}{7} \approx 0.1429 > 0.1414$, so a disc of radius $1/7$ centred at the centre of the cell contains the entire cell, corners included. Equivalently, $\left(\frac{\sqrt{2}}{10}\right)^2 = \frac{2}{100} = 0.02 < \frac{1}{49} \approx 0.0204 = \left(\frac{1}{7}\right)^2$.
*Step 4 -- Conclusion.* Place the glass centred on the cell found in Step 2. It covers the whole cell and therefore at least three ants. Since the arrangement was arbitrary, this is always possible.
Answer: Yes, always. Cut the unit square into a $5 \times 5$ grid of cells of side $1/5$; with 51 ants some cell holds at least 3. A cell of side $1/5$ has circumradius $\sqrt{2}/10 \approx 0.141 < 1/7$, so the glass centred on that cell covers all of it, and hence at least three ants.
Intuition
The numbers are chosen so that pigeonhole lands exactly: $51 = 2 \times 25 + 1$ forces a cell with three ants in a $5 \times 5$ grid, and the radius $1/7$ is just large enough that a disc covers a $1/5$ square, since $\sqrt{2}/5 < 2/7$. Once you see that a region small enough to be covered by the glass must contain three ants, the proof is done. This is the generalised pigeonhole principle: with $n$ objects in $k$ boxes some box holds at least $\lceil n/k \rceil$, a fact that underlies many existence arguments about clustering, collisions, and concentration of points.