Problem 1237

A homogeneous $B C D$ triangular plate's geometric center $S_{A}$ is also the point of application of the resultant force of the elementary gravitational forces acting on it in a dynamic sense. Therefore, we can think of concentrating the plate's mass $m_{A}=k_{1} t_{A}$ at $S_{A}$, where $t_{A}$ is the area of the plate, and the proportionality factor $k_{1}$ represents the "surface density" of the material. Continuing this, it is sufficient to determine the center of mass of a system of four points $S_{X}$, each with mass $t_{X}$, where $X=A, B, C, D$, since in the absence of further information, $k_{1}$ can be assumed to be the same for all four plates.

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Further, let's combine the masses of two points. The common center of mass of the pair of masses at $S_{X}$ and $S_{Y}$ (where $Y \neq X$) is at the point $S_{X Y}$ on the segment $S_{X} S_{Y}$ such that (by the equality of moments) $S_{X Y} S_{X}: S_{X Y} S_{Y}=m_{Y}: m_{X}=t_{Y}: t_{X}$, or, with suitable proportionality factors $k_{2}$ and $k_{3}$, $S_{X Y} S_{X}=k_{2} m_{Y}-k_{3} t_{Y}$. Using the same principle, we can find the desired center of mass $G$ on the segment $S_{A B} S_{C D}$ by choosing $X$ and $Y$ as $A$ and $B$ in one pair of masses.

Furthermore, we use the fact that all four segments $X S_{X}$ pass through the centroid $S$ of the tetrahedron $A B C D$ such that $S X=3 \cdot S S_{X}$, so the tetrahedron $S_{A} S_{B} S_{C} S_{D}$ is obtained from the previous one by a central similarity. (The value of the ratio is $-1 / 3$, but this is not significant.) This implies that the ratio of the areas of any two faces of the new tetrahedron is the same as the ratio of the corresponding original faces $t$, for example (briefly, symbolically):

$$\left(S_{A} S_{B} S_{C}\right):\left(S_{A} S_{B} S_{D}\right)=(A B C):(A B D)=t_{D}: t_{C}$$

Now, let's project the new tetrahedron onto a plane perpendicular to the edge $S_{A} S_{B}$ - in other words, look at it from infinitely far away in this direction - and denote the projections (as apparent images) of the points in the same way as the points themselves. We only see the triangle $S_{A} S_{C} S_{D}$, and consider the ratio of the two sides starting from $S_{A}$ in this triangle. The lengths of $S_{A} S_{C}$ and $S_{A} S_{D}$ are the actual lengths of the altitudes of the triangles $S_{A} S_{B} S_{C}$ and $S_{A} S_{B} S_{D}$, respectively, which are perpendicular to their common side $S_{A} S_{B}$. Therefore, their ratio is the same as the ratio of the areas of these two triangles, which, according to the previous example, is $t_{D}: t_{C}$, and finally $S_{A} S_{C}=k_{4} t_{D}, S_{A} S_{D}=k_{4} t_{C}$, where $k_{4}$ is a suitable proportionality factor.

From the ratio pair $S_{C D} S_{C}: S_{C D} S_{D}=S_{A} S_{C}: S_{A} S_{D}$, it follows that $S_{C D}$ lies on the angle bisector starting from $S_{A}$ in the apparent triangle, and that we also see the desired $G$ on this bisector. However, our lines are the images of planes perpendicular to the plane of the diagram, so $G$ lies on the plane that bisects the dihedral angle between the faces $S_{A} S_{B} S_{C}$ and $S_{A} S_{B} S_{D}$ of the new tetrahedron; thus, the distances of $G$ from these two face planes are equal; with appropriate notation, $r_{D}=r_{C}$.

We apply our result by choosing the first pair of points $S_{X}$ as $A$ and $C$, and then as $B$ and $C$, so $r_{D}=r_{B}$, and then $r_{A}=r_{D}$, so $G$ is equidistant from all four faces of the tetrahedron $S_{X}$. This proves the statement of the problem.

Remarks. 1. The proven statement, without using the centers of mass of the faces, also means that if we place a mass proportional to the area of the opposite face at each vertex of the tetrahedron, the center of mass of the system gives the center of the inscribed sphere of the tetrahedron.

2. A similar center of mass interpretation can be obtained for the centers of the incircles (external tangent circles) of a triangle and the centers of the tangent spheres of a tetrahedron - if we assign a negative mass to one of the sides of the triangle (a repulsive force instead of an attractive force). In the case of a tetrahedron, a total of 8 tangent spheres are possible, as we can assign a negative mass to two faces at the same time, but for example, in the case of a regular tetrahedron, only 5 are created.
3. The proofs can also be carried out using vectors.