I dont understand the problem....So a point inside a convex poliedra satisfies, its distance to the faces of the poliedra is at most $1$? and we have to show the sum is at most $n-2$??? what happens then with the center of a cube with side 2? its distance to the faces is 1, but the sum is $6$....whats wrong?

Yes, he missed some words... How many time we will wait for the correction?...

A friend of mine told me that it said the distances from the point to the vertices of the polyhedron are all $\le 1$ (actually, he told me they were all $1$, but it was over the phone, so maybe I didn't hear well ).

Hmm... Ideas?

I'll solve the problem with equality (i.e. the point in question is the circumcenter of the polyhedron, which has unit circumradius). I think the one with inequality instead of equality can be reduced to this one. For each plane which contains a face of the tetrahedron, consider the cap $\mathcal C_i$ determined by that plane which lies on the opposite side of the plane from the center of the sphere. If the height of $\mathcal C_i$ (i.e. the largest distance from a point on it to the plane containign the face $i$) is $h_i$, then we have to show that $\sum h_i>2$. The area $S_i$ of $\mathcal C_i$ is $2\pi h_i$, so $\sum h_i=\frac{\sum S_i}{2\pi}\ (*)$. However, the caps do not cover the sphere perfectly (i.e. they cover it, but they overlap), so $\sum S_i>4\pi$. this, together with $(*)$, gives the desired result: $\sum h_i>2$.

A beautiful solution, grobber! It works word by word for the general case too (you have never used the assumption that the vertices are on the unit sphere)

I've corrected the statement sorry for the mistake.

Of course he used it fedja, but it can be easily seen that the problem can be reduced to that case.

ikap wrote: Of course he used it fedja. Where? It seems to me that all he needed was that the polyhedron was completely contained inside the sphere of radius 1 centered at the given point inside the polyhedron. Or am I missing something?

fedja, "caps" imply boundaries on the sphere and these are given by the points ON the sphere and their circumscribed circles....

Actually, ikap, I have to agree with fedja here. All you need are a few planes which cut the sphere and determine a cap each. You never use any points on the sphere, just the fact that the planes of the faces cut the sphere so that the caps they determine cover surface of the sphere completely.

oh, yes, of course, i wasn't looking at it from that point of view and I also didn't read word by word your solution grobber, that's why I was always refering in mind to the steps I followed to the solution, allthough this was after the contest.. My fault, sorry.