# Using your Head is Permitted

## November 2011 solution

**Spoiler alert:** The following description contains hints
regarding the solution of last month's riddle. If you still mean to tackle that
other riddle, you may want to hold on before you read this.
Let us begin by recapping from last month's solution:

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.

The smallest dimensions of an axes-parallel parcel that can contain
this parcel are
*d*_{1}|*A*_{1i}| +
*d*_{2}|*A*_{2i}| +
... +
*d*_{n}|*A*_{ni}|,
where *i* is the dimension number.

What we need to prove is that the product of all these is no smaller than the
product of all *d*_{i}.

Let *B* be an *n*-by-*n* matrix whose (*i*,*j*)'th
element is equal to |*A*_{ij}|.
We are asking whether
the product of all elements of *Bd* can be smaller than the product of
all elements of *d*. Note that the product of all elements of *Bd* is
a polynomial in the elements of *d*, all of whose monomials are of degree
*n*.
Furthermore, all these monomials are non-negative. We will show that the
monomial that is associated with the product of all *d*_{i}
is at least 1, therefore showing that the size of this monomial alone (let alone
all of *Bd*) is at least as large as the product of all elements of
*d*.

To show this, note that this coefficient is the permanent of *B*, which is
at least as large as the determinant of *A*. (Both are sums of the same
elements but with different signs. In computing the permanent of *B*, all
signs are positive.) Because *A* is a rotation matrix, its determinant is
known to be 1, so the permanent of *B*, being the coefficient we are
looking for, is at least 1.

Q.E.D.

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