Polyhedron not circumscribed about a sphere Let $P$ be a polyhedron whose faces are colored black and white so that there are more black faces and no two black faces are adjacent. Show that $P$ is not circumscribed about a sphere.
My teacher has conjectured this. I haven't been able to produce a counterexample, neither a solution. Is this claim true? If not a counterexample would be nice. Thanks in advance for helping.
 A: This is exercise 21.3, of Mathematical Omnibus: Thirty Lectures on Classic Mathematics, 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.


