Eleven unit squares inside a larger square

I asked this by Walter Stromquist via email. Here is our mails:

Hello,

I am a mathematician from Finland. I read from the book "Which way did the bicycle go" that you have found a way to pack $11$ squares with side length 1 to a square with side length $3.8772$ or $3.877083$. Do you know which one is correct? I have asked this on Eleven unit squares inside a larger square but got no response. Is there any computation available for the side length?

Best wishes, Jaakko Seppälä

Hello, Jaakko,

Thanks for writing!

I didn't invent the nice packing for 11 squares. I just proved that you can't do better (or as well) with "45-degree packings." The nice packing appeared in a Martin Gardner column, and he attributed it to Walter Trump. Even though my paper came close, I don't think that anyone has proved that Trump's packing is optimal.

Here is one way to do the calculation. I doubt that it is the best way, but it worked for me.

Let's use the diagram of Figure 2 in my paper. (My paper is the one you linked to in Stack Exchange. The figure in Friedman's paper is reflected in a diagonal.)

Call the lower left corner $(0,0)$. Use $(0,y)$ for where a tilted square touches the left edge, and $(x,0)$ for where another tilted square touches the bottom edge. Use $d$ for the tiny vertical distance between the two squares on the right edge.

Use theta for the angle (about 40 degrees) from the lower edge to the lower-right edge of the tilted square.

The the component in the direction $\theta$ of the vector from $(0,y)$ to $(s-2,s-1)$ is $2$. This gives the equation

$[ (s-2,s-1) - (0,y) ] . ( \cos \theta, \sin \theta) = 2$.

Looking at some other vectors gives some more equations:

$[ (s-1,s-2) - ((1,1) ] . (\cos \theta, \sin \theta) = 2$

$[ (s-1, s-3-d) - (x,0) ] . (\cos \theta, \sin \theta) = 1$

$[ (1, s-1) - (s-1, s-2-d) ] . ( -\sin \theta, \cos \theta) = 2$

$[ (0,y) - (x, 0) ] . (-\sin \theta, \cos \theta) = 3$.

Playing the first two equations against each other gives $y = 2$. (That's cute; I never noticed that!) Then it isn't too hard to get equations for $x$ and $s$ in terms of $\theta$, and then and equation for $d$ in terms of $\theta$, $x$, and $s$. That leaves us one equation to test. So we try various values of $\theta$, compute $x$, $s$, and $d$, and then see whether the test equation holds. (Like I said, there must be a better way.)

Anyway, when I do that now I get

$\theta = 40.1819372903297$

$\textrm{side} = 3.877083590022810$.

I did this using Excel, so I wouldn't want to bet too much on the accuracy of the last two digits. But is is pretty clear that Stan Wagon is right, and the number in my paper is wrong. (Stan Wagon is a pretty careful guy, and he is a genius at computation.)

Welcome to the world of square packing! I hope you find many interesting things.

Walter

Hello,

Thank you very much! It was nice to see the computations. So finally I know which one is correct. It would be nice to share this to Stack Exchange but I think there might be some copyright issues.

Jaakko

No copyright issues at this end. Share as you like!