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.



Polyhedron not circumscribed about a sphere
This is exercise 21.3, of [Mathematical Omnibus: Thirty Lectures on Classic Mathematics](https://rads.stackoverflow.com/amzn/click/0821843168), 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.
> 
> 
> 


![](https://ilorentz.org/beenakker/MO/figure213.png)


