From:           Carl Parkes <cparkes@tpgi.com.au>
Date:           Mon, 03 Jun 1996 19:38:17 -0700
Newsgroups:     sci.math
Subject:        Box in a box

this problem has been bugging me.

What is the smalless box (cube) that can be put inside a box (cube) 
touching all the faces? and what is its orentation? 

Carl Parkes

To:             sci.math@wormwood.ICS.UCI.EDU, cparkes@tpgi.com.au
Subject:        Box in a box
Date:           Fri, 02 Aug 1996 14:02:53 -0700
From:           David Eppstein <eppstein@ICS.UCI.EDU>

Several weeks ago, Carl Parkes <cparkes@tpgi.com.au> wrote:
> What is the smallest box (cube) that can be put inside a box (cube) 
> touching all the faces? and what is its orientation? 

The only solutions I have been able to find involve the smaller and
larger boxes sharing a common long diagonal; the best of these solutions
orients the smaller box 180degrees rotated around that diagonal, with
side length 3/5 that of the larger box.  Six of the vertices of the smaller
box are on the faces of the larger box, 3/5 of the way along the face
diagonals; the remaining two vertices are 1/5 and 4/5 of the way along
the long diagonal of the longer box.  (If this is too confusing, see
http://www.ics.uci.edu/~eppstein/junkyard/box-in-box.gif for a picture.)

Are there any solutions in which the two boxes do not share a diagonal?
-- 
David Eppstein		UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu	http://www.ics.uci.edu/~eppstein/

From:           eppstein@ICS.UCI.EDU
To:             cparkes@tpgi.com.au
Subject:        Box in a box
Date:           Mon, 30 Jun 1997 18:12:22 -0700

Remember your question from a year ago, about how to place a small cube
inside a unit cube, so that all the unit cube's faces are touched?
I replied with a set of solutions in which the two cubes share a long
diagonal; in the smallest of these solutions the small cube's size is
3/5 that of the large one.

Anyway, I finally went through a rather ugly proof that these are the
only possible solutions.  The method was to express the requirements
that certain small cube vertices be on the planes of the large cube's
faces in terms of a system of polynomial equations, use Mathematica to
find all families of solutions of these equations, and check that each
solution either has the form I already knew about (with the two cubes
sharing a diagonal) or doesn't work geometrically (for instance,
solutions involving complex numbers).

I've put my calculations online at
http://www.ics.uci.edu/~eppstein/junkyard/box-in-box.nb
if you care to examine them (I'd be just as happy if you didn't;
as I said, they're not pretty).
-- 
David Eppstein		UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu	http://www.ics.uci.edu/~eppstein/

From:           Greg Huber <huber@physics.arizona.edu>
To:             eppstein@ics.uci.edu
Subject:        Re: box in box problem
Date:           Mon, 18 Jan 1999 23:22:06 -0700 (MST)

	Date: Mon, 18 Jan 1999 20:58:07 -0800 (PST)
	From: David Eppstein <eppstein@ics.uci.edu>
	To: Greg Huber <huber@physics.arizona.edu>
	Subject: Re: box in box problem

	Greg Huber writes:

	 > 2) Will your methods work for the d=4 case: 
	 > 
	 >         Find the smallest 4-cube that touches** all 8 cubical faces of
	 >         a unit 4-cube.  What is its side length?
	 > 
	 > ** "touches" here means that some of its vertices lie inside the cubical
	 > faces, and that the small 4-cube lies entirely inside the unit 4-cube.

	The messy symbolic algebra methods could be set up in any dimension,
	but get progressively harder to solve, and were very difficult to solve
	even in 3d.  So those methods would theoretically work but it might be
	very hard to carry them out in practice.

	My 4d visualization isn't good enough for me to guess what the right
	solution is, either, which helped a lot in the 3d case.
	-- 
Yes, same here.  But the 4d case is perhaps more symmetric since you
have 8 cubical faces and 16 vertices, and for a 4d cube there is a very
clear meaning to "every other" vertex.   But I don't think you can get
an optimality argument from this symmetry.   Thank you for your remarks
and postings.  It's a quite lovely class of problems. -- Greg Huber.

From:           David Eppstein <eppstein@ics.uci.edu>
To:             Greg Huber <huber@physics.arizona.edu>
Subject:        Re: box in box problem
Date:           Wed, 20 Jan 1999 09:00:16 -0800 (PST)

Greg Huber writes:
 > Yes, same here.  But the 4d case is perhaps more symmetric since you
 > have 8 cubical faces and 16 vertices, and for a 4d cube there is a very
 > clear meaning to "every other" vertex.   But I don't think you can get
 > an optimality argument from this symmetry.   Thank you for your remarks
 > and postings.  It's a quite lovely class of problems. -- Greg Huber.

OH, DUH.
This idea of taking every other vertex makes it very easy.

The convex hull of every other vertex is a 16-cell (4-dimensional
analogue of an octahedron).  You can also get a 16-cell by taking the
center point of every face of the cube.

So probably the correct answer is to inscribe the small cube so that
every other vertex is at the center of a face of the large cube.

(This is still guesswork, I haven't even verified that the other faces
of the small cube stay inside the large cube.)
-- 
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:           David Eppstein <eppstein@ics.uci.edu>
To:             Greg Huber <huber@physics.arizona.edu>
Subject:        Re: box in box problem
Date:           Wed, 20 Jan 1999 09:03:02 -0800 (PST)

PS would you mind if I added your comments to my page
http://www.ics.uci.edu/~eppstein/junkyard/box-in-box.html?

Also, I don't think I left a pointer from there to
http://www.ics.uci.edu/~eppstein/junkyard/qtvr/
but you might want to check that out -- I have a nice animated gif
(and quicktime vr model) of the box-in-a-box configuration.
-- 
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

http://www.ics.uci.edu/~eppstein/junkyard/box-in-box.html?

From:           David Eppstein <eppstein@ics.uci.edu>
To:             Greg Huber <huber@physics.arizona.edu>
Subject:        Re: box in box problem
Date:           Wed, 20 Jan 1999 09:07:42 -0800 (PST)

PPS it is very easy to see that putting every other vertex of the small
cube at the face centers of the large cube works, and that it is optimal.

Since each opposite pair of vertices of the small cube form a diameter
of the small cube, perpendicular to the large cube's faces, there is no
way to get a pair of vertices touching parallel planes farther apart
than the large cube's faces, and for similar reasons the other 8
vertices of the small cube can't touch or be outside the large cube.
-- 
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:           David Eppstein <eppstein@ics.uci.edu>
To:             huber@physics.arizona.edu
Subject:        explicit coordinates
Date:           Wed, 20 Jan 1999 10:26:52 -0800 (PST)

Let the large box's coordinates be (+-1, +-1, +-1, +-1).
Then the eight small box vertices that touch the large box's faces
have coordinates that are permutations of (0, 0, 0, +-1).
To get the other eight vertices, take every other corner of
the hypercube (+-1/2, +-1/2, +-1/2, +-1/2) (a shrunken copy of the
large cube). The small cube has half the side length of the large one.

Actually there are two different ways of inscribing the small cube in
the large one, depending on which set of every other vertex of the
shrunken cube you take.  These two small inscribed cubes, together with
the cube (+-1/2, +-1/2, +-1/2, +-1/2) itself, form a very nice complex
of three cubes in which each pair of two cubes shares the eight vertices
of a 16-cell.  The whole set of 24 vertices of this complex forms
another regular 4-polytope, the 24-cell, which is self-dual and has 24
octahedral faces.
-- 
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

From:           Greg Huber <huber@physics.arizona.edu>
To:             eppstein@ics.uci.edu
Subject:        Re:  explicit coordinates
Date:           Wed, 20 Jan 1999 11:56:16 -0700 (MST)

David -- Thanks for your comments and observations.  Yes, the one
you talk about in your messages is the one I had in mind, but I 
did not at first see your argument for optimality -- very nice! 

Yes, please add my problem, etc and your solution to the web page.
Thank you for the additional link -- I have not really explored your
web page beyond the text that I mentioned earlier -- I'm sure there
a lot there that I would enjoy.   Best -- Greg.