Eleven unit squares inside a larger square

What is the smallest square which contains 11 non-overlapping (except boundary) unit squares?

This question is open but I would like to know a method to verify the best known answer at the moment.

I'm reading paper at here . In figure 2, there has been said that the best known packing has side length about 3.8772 and tilt angle about 40.182 degrees. How can I verify those values, as another source (Which Way did the Bicycle Go page 105) says that the side length is about 3.877083? I think I have to denote the tilt angle by $\alpha$, a side of the square by $x$ and form two equations but I don't see how to make those equations.

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Smallest in what sense? You can trivially put them all in a $1\times 11$ rectangle, but the perimeter of this (for example) would be larger than that of the square given in the pdf... – Paul VanKoughnett Nov 3 '10 at 0:56
You may want to edit your question to say "square" where it says "rectangle" now, to agree with the question title and the linked paper. – Rahul Nov 3 '10 at 1:16
Oops. My blunder. Fixed. – Jaska Nov 3 '10 at 1:41
Could this be verified by computer? Namely assume that the center points of unit square are rationals $a_i/b_i$ where $\max\{a_i,b_i\}<k$, the angle $\alpha_i$ between side of square number $i$ is $\alpha_i$ taking discrete values and $k$ is fixed. Then there is finite numbers to be checked so some algorithm might give and approximation whether side lenght > 3.8771 or not. Then check that error is small enough. Other algorithm that comes to my mind is to solve the following problem effectively. Given an $n$-gon $A$ with given end point of edges. Can we put an unit square inside $A$? – user45685 Oct 22 '12 at 21:48

It's a pity this example isn't really describe anywhere. Assuming that there's a triple intersection at $(1,2)$, we get the equation

$$k - 3 - \sin\alpha - \cot\alpha\cdot(2 + 2\cos\alpha - k) = 0$$

Presumably you can get more equations by assuming that the tilted squares touch each other and corners of the straight squares, but these seem not to be satisfied by the given $k,\alpha$.

How to find more equations: the tilted squares come in three groups (two, two and one), and all seem to be just touching each other. For each of the groups, one vertex has one known coordinate (the middle group might have a known vertex at $(1,2)$, but I don't know if that's true). You can get equations from the fact that these groups touch certain corners of the straight squares, and from the (apparent) fact that they just touch each other.

The simplest way to get these equations is to define vectors $s,t$ for the sides of the tilted square (they can be expressed in terms of $\alpha$), and now all these touchings amount to pairs of points $p,q$ which are parallel to either $s$ or $t$. So $p-q$ is proportional to $s$ (or $t$), and we get an equation.

Unfortunately, pursuing this route, I get conflicting equations (conflicting in the sense that when I substitute $k$ and $\alpha$, they are contradictory). Maybe you can try and tell us if it works out for you.

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Check out the squares in squares page in Erich's Packing Center, and [http://www2.stetson.edu/~efriedma/papers/squares/squares.html#figure6](his survey paper he links to).

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