# Is the lower bound proof in this paper correct?

In this paper on "Circle Packing for Origami Design Is Hard" by Erik D. Demaine, Sandor P. Fekete, Robert J. Lang, on page 15, figure 13, they claim that the side length of the smallest square that encloses two circles of area 1/2 each is 1.471299. By my calculations I am getting side length 1.362 and area 1.855. Have I made a mistake or is there a mistake in the paper?

• I tried to "reverse engineer" and figure out where the number 1.471299 comes from, but I had no luck. – Robin Kothari Sep 1 '10 at 0:07
• @Robin: Look at possible closed forms: wolframalpha.com/input/?i=1.471299 – Casebash Sep 1 '10 at 5:53
• @Casebash: Thanks. I didn't know Wolfram Alpha could do that. – Robin Kothari Sep 1 '10 at 6:58
• @Robin: Interesting. Although I can't immediately see any relevance of any of these expressions to the problem at hand. – Vinayak Pathak Sep 2 '10 at 23:27

I get the same answer as you, $\frac{1+\sqrt{2}}{\sqrt{\pi}}$.
If I place two unit disks inside a larger square, their centers are one unit away from the left and right sides, and $\sqrt{2}$ units apart from each other in $x$-coordinates, so the side of the larger square is $2+\sqrt 2$. The unit disks have area $\pi$, and we want area $\frac{1}{2}$, so dividing the side length and the disk radius by $\sqrt{2\pi}$ gives the result.