'Twas the night before Christmas and throughout the net,
Not a question was posed, at least---not yet.
When what to my horror I suddenly realized:
One present was not wrapped, a present most prized.
For a juggler of the future, three balls all the same.
But how best to foil-wrap, within a tight frame?
> ***Q1***. What is the smallest square that can wrap three unit-radius balls, without cutting the square?
To *wrap* means to completely cover their convex hull.
I mention "foil" above because one may want to crinkle the
wrapping over sphere caps, analogous to how
[Mozartkugeln](https://en.wikipedia.org/wiki/Mozartkugel) are wrapped.<sup>1</sup>
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[![Balls3][1]][1]
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Likely easier is this question, which may only require rough bounds:
> ***Q2***. Which of the two configurations shown above is easier to wrap,
easier in the sense that a smaller square suffices?
<hr />
<sup>1</sup>
The square of diagonal $2 \pi$ is the smallest square that wraps a unit-radius sphere.
Demaine, Erik D., Martin L. Demaine, John Iacono, and Stefan Langerman.
"Wrapping spheres with flat paper." *Computational Geometry* 42, no. 8 (2009): 748-757.
[Journal link](https://www.sciencedirect.com/science/article/pii/S0925772109000182).
<hr />
*Added*. I thought I would compute the surface areas of the
convex hulls of the two configurations.
For the linear configuration,
<hr />
[![Linear3Balls][2]][2]
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I compute
$$4 \cdot 2 \pi + 4 \pi = 12 \pi \approx 37.7 \;.$$
For the triangular configuration,
<hr />
[![Tri3Balls][3]][3]
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I compute $4 \pi$ for the three $\frac{1}{3}$ spheres,
three times $2 \cdot \pi$ for the cylinder pieces,
and two flat $\sqrt{3}$ triangles:
$$4 \pi + 6 \pi + 2 \sqrt{3} = 10 \pi + 2 \sqrt{3} \approx 34.9 \;.$$
Of course, this does not address which is easier
to wrap with a square.
[1]: https://i.sstatic.net/dGfCz.jpg
[2]: https://i.sstatic.net/uHXvQ.jpg
[3]: https://i.sstatic.net/PZFDY.jpg