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I know how to pack $5$ unit squares in a square of side length $2+\frac{\sqrt{2}}{2}$. Is there an $\varepsilon>0$ such that there exists a packing of $9$ unit cubes in a cube of side length $3-\varepsilon$?

(Inspired by this question.)

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@Dustin, I see only a very weak analogy between the 2d and 3d situations described in your question. What am I missing? – Włodzimierz Holsztyński May 19 '13 at 0:16
When I asked the question, I was assuming the 9th cube had to be centered in the center of the $3−\varepsilon$ cube, but I couldn't tell if there was a way to rotate in a way that avoided the corner cubes as there clearly is in the 2-dimensional analog. – Dustin G. Mixon May 19 '13 at 0:19
If you center the $9$th cube, you can rotate it $45$ degrees about an axis parallel to an edge, and then when you project perpendicular to that edge you get the $2$-dimensional picture. – Douglas Zare May 19 '13 at 21:04
Yeah, I realized that after I saw the link in Ricky's answer. Considering I already understood the 2-dimensional setting, I definitely had a brain fart! – Dustin G. Mixon May 19 '13 at 22:25
up vote 8 down vote accepted

Yes, since one can pack 10 unit cubes in a cube of side length $\:\:2+\left(\hspace{-0.023 in}\frac12\hspace{-0.05 in}\cdot\hspace{-0.03 in}\sqrt2\right) \;\;$.

See $\:$ .

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Yeah, good point. But 14 cubes?? Wow! – Dustin G. Mixon May 18 '13 at 20:05
As far as I know, 15 might be possible. $\:$ – Ricky Demer May 18 '13 at 20:07

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