27 squares of side $1$ inside a circle of radius $2$: Show that three squares share an interior point.

Question

By placing 27 squares with side $1$ in a circle with radius $2$, such that each square is entirely inside the circle, prove that there exists a point that 3 squares share (inside of their area).

I thought about dividing the circle into 9 equal areas and using the pigeonhole principle, but got stuck there. Any hint will help.

Answer 1

To get the desired conclusion, $26$ unit squares will suffice.

Let $D$ denote the disk of radius $2$, and let $C$ be the subregion of $D$ covered by some placement of $26$ unit squares within $D$.

Suppose every point of $D$ is contained in at most two of the unit squares.

Our goal is to derive a contradiction.

Let $X_i$ denote the $i$-th unit square.

For each $i$, let

• $A_i$ be the subset of $X_i$ which does not intersect any of the other squares.$\\[4pt]$
• $B_i = X_i{\setminus}A_i$.

For a given set $S$, let $[S]$ denote the area of $S$. \begin{align*} \text{Then}\;\; 4\pi&=[D]\\[4pt] &\ge [C]\\[4pt] &= \sum_{i=1}^{26}[A_i] + \left({\small{\frac{1}{2}}}\right) \sum_{i=1}^{26}[B_i]\\[4pt] &\ge \left({\small{\frac{1}{2}}}\right) \sum_{i=1}^{26}[A_i] + \left({\small{\frac{1}{2}}}\right) \sum_{i=1}^{26}[B_i]\\[4pt] &= \left({\small{\frac{1}{2}}}\right) \sum_{i=1}^{26}\left([A_i] + [B_i]\right)\\[4pt] &= \left({\small{\frac{1}{2}}}\right) \sum_{i=1}^{26}\,(1)\\[4pt] &= \left({\small{\frac{1}{2}}}\right)(26)\\[4pt] &= 13\\[4pt] \end{align*} contradiction.

This proves the claim.

Note:

Squares are an awkward shape for covering a large subregion of a disk, so I suspect the actual maximum number of unit squares which can be placed in a disk of radius $2$, such that each point of the disk is covered by at most two of the unit squares, is a lot less than $25$.

Answer 2

Lets denote by $A$ the area that the squares occupy.

Of course $A\leq4\pi$(the area of the circle). Let's also suppose that there are no 3 squares that share a point.

Therefor only $2$ squares can intersect, i.e. $A>\frac{27}{2}$, so $$8\pi>27$$ which is false

Answer 3

@razvanelda: With your reasoning wouldn’t 26 squares be sufficient in order to force 3 squares to have a common interior point since $8\pi>26$ is also false?