I was wondering a question given in Eleven unit squares inside a larger square and topology.
The problem claims that 11 squares of side length 1 can be packed to a square with side length 3.87709 such that no squares of side length 1 have common interior points.
How do the existence of such a minimal square can be proven? I mean, what kind of argument proves that a square of some side length $s$ contains some 11 squares of side length 1 configuration with no interior points but for every $\epsilon>0$, no square of side length $s-\epsilon$ contains the same configuration? And how one can prove that in general $n>1$ unit squares with side length 1 and no interior points, there is such a configuration that fits inside some larger square but do not fit in any smaller square?
Let $A$ be the set of all $x\in\Bbb R$ such that $11$ unit squares with disjoint interiors can fit in a square of side $x$. Then $A$ is bounded below; we can't fit $11$ unit squares in a square of side $1/2$, say. Also, $A$ is non-empty; we can easily fit $11$ unit squares in a square of side $12$.
So by the completeness axiom for the reals, $A$ has a greatest lower bound $s$. What this means is that for any $\epsilon>0$, we can fit $11$ unit squares in a square of side $s+\epsilon$; and we can't fit $11$ unit squares in a square of side $s-\epsilon$.
So far so good. But your question explicitly asks whether we can fit $11$ unit squares in a square of side exactly equal to $s$; and the completeness axiom doesn't help us here. At first glance it is plausible that there may be a configuration for every square of side greater than $s$, but not for the square of side $s$.
To see that this can't happen, we need compactness. Consider a unit square constrained so that its centre lies inside the square $[0,s]\times[0,s]$. We can represent its position and orientation by an element of $[0,s]^2\times S$, where $S$ is the unit circle. So we can represent the positions and orientations of $11$ such squares by an element of $\mathscr U=[0,s]^{22}\times S^{11}$.
Now we define a function $F:\mathscr U\to\Bbb R$ as the sum of the overlaps and the extrusions of a configuration in $\mathscr U$. Here the overlap of two unit squares is the area of their intersection; and the extrusion of a unit square is the area of the square that lies outside the larger square $[0,s]\times[0,s]$. $F$ is the sum of the overlaps over all pairs of unit squares, plus the sum of the extrusions of all unit squares.
OK, so what we are looking for is a configuration $u\in\mathscr U$ with $F(u)=0$. How do we know that such a configuration exists? We use two ingredients:
$[0,s]$ and the unit circle $S$ are both compact; so $\mathscr U=[0,s]^{22}\times S^{11}$ is also compact.
And we know that for any $n\in\Bbb N$, there exists a configuration $u_n$ of unit squares that have disjoint interiors (so that the sum of their overlaps is $0$), and which all fit in the square $[0,s+\frac{1}{n}]\times [0,s+\frac{1}{n}]$, so that the sum of their extrusions is at most $11\left(\frac{2s}{n}+\frac{1}{n^2}\right)$.
Thus $F(u_n)\le 11\left(\frac{2s}{n}+\frac{1}{n^2}\right)$, which tends to $0$ as $n\to\infty$.
And $F$ is clearly continuous. So by the compactness of $\mathscr U$, the sequence $(u_n)$ has a convergent subsequence $(u_{n_i})$ with $F(\lim_{i\to\infty}u_{n_i})=\lim_{i\to\infty}F(u_{n_i})=0$. QED.
I realise that this is all a bit clunky, but I think it hangs together. I would love to see a more elegant solution, if anybody can think of one.