A tetrahedron is cut by planes parallel to its faces and tangent to the inscribed sphere, resulting in four smaller tetrahedra. What is the radius of the sphere inscribed in the original tetrahedron if the radii of the spheres inscribed in the smaller tetrahedra are 9, 12, 36, and 39 units long?
I. Solution. Let the vertices of the tetrahedron be denoted by $A, B, C, D$, the lengths of the altitudes by $m_{A}, m_{B}, m_{C}, m_{D}$, the areas of the corresponding faces by $T_{A}, T_{B}, T_{C}, T_{D}$, the volume of the tetrahedron by $V$, the radius of the inscribed sphere by $\rho$, and the radii of the spheres inscribed in the cut-off small tetrahedra by $\rho_{A}, \rho_{B}, \rho_{C}, \rho_{D}$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_c5f0aae59669b6bdbb81g-1.jpg?height=539&width=548&top_left_y=305&top_left_x=775)

The cut-off small tetrahedra are similar to the original tetrahedron (Figure 1), since the theorem of parallel chords implies that the central dilation from the common vertex of the two tetrahedra, with the appropriate ratio, transforms the small tetrahedron into the large one. Therefore, the ratios of corresponding segments in the tetrahedra are equal. The altitude corresponding to the common vertex of the small tetrahedra is equal to the difference between the corresponding altitude of the large tetrahedron and the diameter of the inscribed sphere. Writing the ratios of the altitudes and the radius of the inscribed sphere:

$$
\frac{\rho_{A}}{\rho}=\frac{m_{A}-2 \rho}{m_{A}}, \quad \frac{\rho_{B}}{\rho}=\frac{m_{B}-2 \rho}{m_{B}}, \quad \frac{\rho_{C}}{\rho}=\frac{m_{C}-2 \rho}{m_{C}}, \quad \frac{\rho_{D}}{\rho}=\frac{m_{D}-2 \rho}{m_{D}}
$$

Rearranging and adding these equations, we get:

$$
\rho_{A}+\rho_{B}+\rho_{C}+\rho_{D}=4 \rho-2 \rho^{2}\left(\frac{1}{m_{A}}+\frac{1}{m_{B}}+\frac{1}{m_{C}}+\frac{1}{m_{D}}\right)
$$

We know that the volume of the tetrahedron satisfies:

$$
3 V=T_{A} \cdot m_{A}=T_{B} \cdot m_{B}=T_{C} \cdot m_{C}=T_{D} \cdot m_{D}
$$

From this, we have:

$$
\frac{1}{m_{A}}+\frac{1}{m_{B}}+\frac{1}{m_{C}}+\frac{1}{m_{D}}=\frac{T_{A}+T_{B}+T_{C}+T_{D}}{3 V}
$$

Substituting this into (1):

$$
\rho_{A}+\rho_{B}+\rho_{C}+\rho_{D}=4 \rho-2 \rho \cdot \frac{\rho\left(T_{A}+T_{B}+T_{C}+T_{D}\right)}{3 V}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c5f0aae59669b6bdbb81g-1.jpg?height=547&width=507&top_left_y=1911&top_left_x=793)

If $O$ is the center of the inscribed sphere of the tetrahedron, then the altitude of the tetrahedron $O A B C$ corresponding to $O$ is $\rho$ (Figure 2), so its volume is $\frac{T_{A}}{3} \rho$. Adding four such equalities, we get $3 V=\rho T_{A}+\rho T_{B}+\rho T_{C}+\rho T_{D}$. Substituting this into (2):

$$
\rho_{A}+\rho_{B}+\rho_{C}+\rho_{D}=4 \rho-2 \rho=2 \rho
$$

Using the values given in the problem:

$$
\rho=\frac{1}{2}(9+12+36+39)=48 \text { units. }
$$

We also need to check whether there exists a tetrahedron for which the radii of the spheres inscribed in the cut-off small tetrahedra are 9, 12, 36, and 39 units. It can be shown that a tetrahedron with three mutually perpendicular edges (emanating from a common vertex) of lengths 128, 384, and 512 units, respectively, has exactly these properties.

II. Solution. Using the notation from the first solution, and denoting the radii of the exspheres of the original tetrahedron by $\rho_{a}, \rho_{b}, \rho_{c}, \rho_{d}$ respectively.

It is known (see, for example, Collection of Geometric Problems I., Problem 2006/a), that in this case:

$$
\frac{2}{\rho}=\frac{1}{\rho_{a}}+\frac{1}{\rho_{b}}+\frac{1}{\rho_{c}}+\frac{1}{\rho_{d}}
$$

The exsphere of the cut-off tetrahedron is precisely the inscribed sphere of the original tetrahedron, so due to the similarity established in the first solution:

$$
\frac{\rho_{A}}{\rho}=\frac{\rho}{\rho_{a}}, \quad \frac{\rho_{B}}{\rho}=\frac{\rho}{\rho_{b}}, \quad \frac{\rho_{C}}{\rho}=\frac{\rho}{\rho_{c}}, \quad \frac{\rho_{D}}{\rho}=\frac{\rho}{\rho_{d}}
$$

Solving for $\rho_{a}, \rho_{b}, \rho_{c}$, and $\rho_{d}$ and substituting into (3), we get:

$$
\frac{2}{\rho}=\frac{\rho_{A}+\rho_{B}+\rho_{C}+\rho_{D}}{\rho^{2}}
$$

which simplifies to:

$$
\rho=\frac{\rho_{A}+\rho_{B}+\rho_{C}+\rho_{D}}{2}
$$

In our case, the radius of the inscribed sphere is $\frac{9+12+36+39}{2}=48$ units.

Based on the work of Lóránt Kassai (Bp., Fazekas M. Gyak. Gimn., I. o. t.)