<p>I get the same answer as you, $\frac{1+\sqrt{2}}{\sqrt{\pi}}$.</p>

<p>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.</p>

CC BY-SA 2.5
2010-08-31T20:52:00.860Z
2010-08-31T20:52:00.860Z
David Eppstein
<p>In <a href="http://arxiv.org/abs/1008.1224" rel="noreferrer">this paper</a> 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?</p>

CC BY-SA 2.5
2010-09-01T00:07:31.140Z
I tried to "reverse engineer" and figure out where the number 1.471299 comes from, but I had no luck.
Robin Kothari
CC BY-SA 2.5
2010-09-01T05:53:55.610Z
@Robin: Look at possible closed forms: http://www.wolframalpha.com/input/?i=1.471299
Casebash
CC BY-SA 2.5
2010-09-01T06:58:57.183Z
@Casebash: Thanks. I didn't know Wolfram Alpha could do that.
Robin Kothari
CC BY-SA 2.5
2010-09-02T23:27:52.267Z
@Robin: Interesting. Although I can't immediately see any relevance of any of these expressions to the problem at hand.
Vinayak Pathak
<cg.comp-geom>
Is the lower bound proof in this paper correct?
Vinayak Pathak
# Is the lower bound proof in this paper correct?

(Asked by: username_0 on 2010-08-31)

In [this paper](http://arxiv.org/abs/1008.1224) 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?

**Tags:** cg.comp-geom

> **Comments:**
> **username_1**: I tried to "reverse engineer" and figure out where the number 1.471299 comes from, but I had no luck.  
> **username_2**: @Robin: Look at possible closed forms: http://www.wolframalpha.com/input/?i=1.471299  
> **username_1**: @username_2: Thanks. I didn't know Wolfram Alpha could do that.  
> **username_0**: @Robin: Interesting. Although I can't immediately see any relevance of any of these expressions to the problem at hand.  

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## 1 answers

### Answer by username_3 on 2010-08-31. Score: 18 | Accepted answer

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.


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