Newsgroups: rec.puzzles
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Date: 8 May 92 14:49:12 GMT
Subject: Re: Hypercube puzzle

schni...@cs.ucf.edu (Mark Schnitzius) wrote:

|> As you may or may not know, the largest area that
|> a 1x1x1 cube can have, when projected onto a
|> 2 dimensional plane, is sqrt(2) square units

When I saw this claim, I thought it was quite strange that the maximum
would be achieved in such a non-symmetric orientation, but I didn't
get around to checking it.

w...@math.canterbury.ac.nz (Bill Taylor) replied:

> A cube can project a bigger area than that, as follows...
>         _________
>       / \        \
>      /   \        \
>     /     \________\
>     \     /        /
>      \   /        /
>       \ /        /
>        ~--------'
> ... giving A-max = sqrt(3)

After seeing this, I verified that you can get the maximum area to be
sqrt(3), and it's in exactly the configuration I expected.

> Further calculation shows that, in the max configuration given above, the
> projected cross-section is in fact a regular hexagon, side sqrt(2/3)

Furthermore, the other two vertices map to the center of the hexagon.

There was a discussion about the ``Halved Hypercube'' last month, and
this is just the three-dimensional case.  I suspect the answer to
Mark's question is that you maximize the projected hyper-area by
projecting two antipodal vertices together, yielding the same answer
as the halved hypercube.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil