Prove that we cannot inscribe sphere in polyhedron.

Each face of a polyhedron is colored with red or blue such that each two faces with a common edge are colored differently. Suppose that the blue area is larger than the red. Prove that we cannot inscribe a sphere in the polyhedron.

Firstly I thought that this one has to do something with Euler formula for the planar graphs. But I have no idea how to start. Any help?

• Do you mean convex (3D) polyhedron? and is the inscribe sphere required to touch all the faces? – Ehsan M. Kermani Oct 18 '17 at 1:04
• @Ehsan M. Kermani: Yes. – Aqua Oct 18 '17 at 8:51
• If three faces meet ay a vertex, how do you color the third face? – Aretino Oct 18 '17 at 14:53
• @Aretino you don't; in order to meet the requirements of the problem, the polyhedron must have even degree vertices only, so that's what we're stuck with. – Dan Uznanski Oct 27 '17 at 2:56

pick a center $C$, and break up the polyhedron based on its edges, so that each edge is associated with the two triangles of its faces that it makes along with the point where a sphere with that center would meet each face.

I say that, if a sphere can be inscribed within the polyhedron from this center, the two triangles for each edge would be equal.

The plane that is normal to the edge and passes through the center also passes through the sphere-face meeting point $P$, and these two points and the point where the edge and plane meet $N$ make a right triangle with $P$ as the vertex. This gives us a couple of things: no matter what, $P$ is on a circle with $NC$ as a diameter, and $NP$ is the height of the edge's triangle with respect to the edge.

In order for a sphere to be inscribed, both $P$ for any edge must be the same distance from $C$, so the two $P$ are on a circle around $C$. Now we have two circles, and two circles can only intersect in two points, both of which will be the same distance also from $N$ (and indeed from any point along $NC$). So both $NP$ are the same length, and so the height of both triangles are the same, and since they both use the same base, their areas must be equal!

But here, every edge has two colors of face attached, and we know that these two colors add up to different amounts. How do we get unequal things by adding equal values? We cannot! Thus, no matter what center we choose, we are doomed: there must be an edge for which the two associated triangles are of different areas.

• Are not every two triangles which are associated to edge even congruent? – Aqua Oct 18 '17 at 8:42
• If we have edge $AB$ for the faces $F$ and $F'$ whic circle touches at $T$ and $T'$, then $\triangle ATC\cong \triangle AT'C$ (ssa) because they share the same side $AC$ and $TC=TC'$ (by assumption). So we have $AT=AT'$ and the same we can say for $BT=BT'$. Thus $\triangle ATB\cong \triangle AT'B$ (sss). – Aqua Oct 18 '17 at 8:50
• Congruent indeed, if our center is indeed the center of an inscribed sphere. – Dan Uznanski Oct 18 '17 at 8:57