Two tetrahedra are congruent given a certain condition

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.

This is a comment too long to fit in the usual comment format.

One important reduction that we can make is to reduce this three-dimensional problem to a planar one. Indeed, let ${\cal F }=ABC$ be any face of one of the tetrahedra, and let $H$ be the orthogonal projection of $O$ on $ABC$. Let $d=OH$ (this is the distance of $O$ to the plane $(ABC)$). Denote by ${S'}_r$ the disk of center $H$ and radius $\sqrt{r^2-d^2}$. Then the intersection $S_r \cap {\cal F}$ in ${\mathbb R}^3$ is the exactly the same thing as the intersection ${S'}_r \cap {\cal F}$ in the plane $(ABC)$ (if $r \geq d$).

This leads us to the following definition. Let $\cal P$ be a plane, and let $H,A,B,C\in {\cal P}$ such that $ABC$ is a nondegenerate triangle. For $r>0$, denote by $f(H,ABC,r)$ the area of the intersection of the convex hull of $A,B,C$ with the disk of center $H$ and radius $r$.

Then, if we return to our initial situation where ${\cal T}$ is the tetrahedral convex hull of $A,B,C$ and $D$, and $O$ is inside $\cal T$. Enumerate the faces of $\cal T$ in some order : $F_1,F_2,F_3,F_4$. Denote by $H_k$ the orthogonal projection of $O$ on the plane spanned by $F_k$. Then we have

$$ (*) |S_r \cap {\cal T}|=\sum_{k=1}^{4} f(H_k,F_k,\sqrt{r^2-OH_k^2}) $$

A sufficient condition for the answer to the initial problem to be yes is the following "purely planar"statement : if inside a plane we take one point $\Omega$ and several noncongruent triangles $(A_kB_kC_k)_{1 \leq k \leq m}$ then the functions $ r \mapsto f(\Omega,A_kB_kC_k,r)$ (for $1 \leq k \leq m$) are linearly independent. I am almost certain this last condition is true.

Clearly, a good way to understand the functions $ r \mapsto f(\Omega,ABC,r) $ is to look at invariants such as the distances of $\Omega$ to the edges. But there are other, more complicated invariants, such as the distance $r_0$ between $\Omega$ and the furthest point among $A,B$, and $C$ (the function $f$ is constant for $r\geq r_0$ and $r_0$ is minimal with this property).

It is not clear to me that the distances between $\Omega$ and the vertices (the distances other than $r_0$ I mean) can be reconstructed from $f$. Perhaps someone else can fill the gap here.