This is exercise 21.3, of <A HREF="http://www.amazon.com/dp/0821843168/ref=rdr_ext_tmb">Mathematical Omnibus: Thirty Lectures on Classic Mathematics</A>, by D.B. Fuks and Serge Tabachnikov. The solution is given on page 448:

> Assume that $P$ is circumscribed. Consider a face $A_1$, $A_2$,...
> $A_n$ and let $O$ be its tangency point with the sphere. Clearly, the
> sum of angles $A_i O A_{i+1}$ is $2\pi$. We shall sum up al these
> angles over all faces, taking the angles in the white faces with
> positive signs and the angles in the black faces with negative signs.
> Since there are more black faces, this sum, $\Sigma$, is negative.
> 
> On the other hand, consider two adjacent faces with a common edge
> $AB$; see figure; The angles $AOB$ and $AO'B$ are equal. Indeed,
> revolve the plane $AOB$ about the line $AB$ (as if it were a hinge)
> until it coincides with the plane $AO'B$. This rotation takes point
> $O$ to $O'$ and hence the triangles $AOB$ and $AO'B$ are congruent.
> 
> There are two kinds of adjacent faces; black-white and white-white;
> the former's contribution to $\Sigma$ cancels, and the latter
> contribute a positive number. Thus $\Sigma\geq 0$, a contradiction.

<IMG SRC="http://ilorentz.org/beenakker/MO/figure213.png">