Note that if $XA$ and $XB$ are tangent to some Sphere at $A,B$ respectively. Then $XA = XB$ (One simply see this by considering the plane determined by $XAB$, which intersects the spheres in a circle, whence the result follows clearly from plane Geometry). We note also that any convex polygon can be partitioned, by a point inside it into several triangles having that point common, that's for a polygon $A_1 A_2 \cdots A_n $, if $P$ is a point in its interior, we mean the partition composed of the triangles $A_i P A_{i+1}$. Clearly, the sum of areas of such triangles equals that of the polygon. Now we partition each face of the polyhedral in that manner, with the interior point be the associated point of tangency. Using the first observation any two of such triangles in adjacent faces sharing the same edge are congruent, hence have the same area. As each black face has all its adjacent faces white, we map each black triangle, to the white triangle which shares the face's edge in the adjacent face. Now we prove that this mapping is injective. But that can be seen as every edge of a white face is shared with at most one black face, so each white triangle is the image of at most one black triangle. As every triangle is mapped to a one with the same area, the result follows