This question is inspired by a Miklos Schweitzer problem, namely
Problem 9./2007 Let $A$ and $B$ be two triangles in the plane such that the interior of both triangles contains the origin, and for each circle $C_r$ centred at the origin $|C_r \cap A|=|C_r \cap B|$ (where $|\cdot|$ is the arclength measure). Prove that $A,B$ are congruent. Does this statement remain true if the origin lies on the border of $A$ or $B$?
This problem can be solved relatively easy proving that the distances from the origin to the edges and the vertices of the two triangles are the same for $A$ and $B$. To do this, consider the circle $C_r$ "growing" until it first touches one side of $A$. If it doesn't touch a side of $B$, making $r$ a little bigger, we get a contradiction, since $C_r$ is still inside of $B$, but a part of it is outside $A$ now. Therefore, the distance from $O$ to the nearest side of $A,B$ is the same. Now consider the next side and so on.
I was wondering how to apply the same reasoning to the following generalization of the problem:
Consider two tetrahedra $A,B$ which both contain the origin in the interior, with the property that for any sphere $S_r$ centred at the origin we have that $|S_r \cap A|=|S_r \cap B|$ (where $|\cdot |$ is the area measure in $\Bbb{R}^3$). Prove that the two tetrahedra are congruent.
For the triangle, it is enough to know that the distances from the origin to the sides and vertices are equal to prove that they are congruent. For the tetrahedron I feel I need something more.