# Using your Head is Permitted

## October 2011 solution

The answer is that you cannot. Not in three dimensions and not in any other
number of dimensions.
Let *d* be an *n*-dimensional vector denoting the dimensions of the
parcel. Suppose we
wish to pack it inside a larger parcel by placing it at some odd angle. We
describe this angle by an *n*-by-*n* matrix, *A*, whose rows are
the unit vectors in the directions of the main axes of the parcel.

What are the smallest dimensions of an axes-parallel parcel that can contain
this parcel?

If *X* and *Y* are two diametrically opposite corners of the parcel,
the vector connecting *X* and *Y* is
±*d*_{1}*A*_{1}
±*d*_{2}*A*_{2}
...
±*d*_{n}*A*_{n}.

By picking all 2^{n} possible *X*-*Y* pairs, any
assignment of signs to the previous equation can be attained.
For the larger parcel to contain all these vectors, it must be at least as
large as the largest of these in every dimension. Therefore, the *i*'th
dimension of the parcel must be at least
*d*_{1}|*A*_{1i}| +
*d*_{2}|*A*_{2i}| +
... +
*d*_{n}|*A*_{ni}|,
which, summed together, yields
*d*_{1}(Σ_{i} |*A*_{1i}|) +
*d*_{2}(Σ_{i} |*A*_{2i}|) +
... +
*d*_{n}(Σ_{i} |*A*_{ni}|).

We will prove that for any *j*,
Σ_{i} |*A*_{ji}| ≥ 1.

To do this, we recall that the transpose of a rotation matrix is also a
rotation matrix. (This is a fact that appeared previously on this site. It can
be seen immediately when considering that a rotation matrix can be represented
as the product of matrices that represent axis-parallel rotations. The transpose
can therefore be represented as a product of transposed axis-parallel rotation
matrices, which can easily be verified to be axis-parallel rotation matrices
themselves.)

We therefore know that Σ_{i} *A*_{ji}^{2}=1. Compare this with
(Σ_{i} |*A*_{ji}|)^{2}, and
the solution is immediate.

Now for credits:

This is a puzzle of Russian origins. It made an appearance in the
International Mathematics Tournament of the Towns (Problem 5, Fall 1998,
Senior A-Level Paper). However, it is older than this. I first saw it with a
different solution in a Peter Winkler collection.

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