Problem 11.8. Inside the tetrahedron $A B C D$, points $X$ and $Y$ are given. The distances from point $X$ to the faces $A B C, A B D, A C D, B C D$ are $14, 11, 29, 8$ respectively. And the distances from point $Y$ to the faces $A B C, A B D, A C D, B C D$ are $15, 13, 25, 11$ respectively. Find the radius of the inscribed sphere of the tetrahedron $A B C D$.
Answer: 17.

Solution. Consider a point $Z$ lying on the ray $XY$ such that $XY: YZ = 1: 2$. We will prove that this point is the center of the inscribed sphere of the tetrahedron.

Drop perpendiculars $X_{\alpha}, Y_{\alpha}, Z_{\alpha}$ from points $X, Y, Z$ to the plane $\alpha$ - obviously, they will lie in the same plane perpendicular to $\alpha$ (Fig. 9a). Also, draw a line through $X$ parallel to $X_{\alpha} Y_{\alpha}$, and denote its intersections with the lines $Y Y_{\alpha}$ and $Z Z_{\alpha}$ as $Y^{\prime}$ and $Z^{\prime}$, respectively (Fig. 9b).

Since triangles $X Y Y^{\prime}$ and $X Z Z^{\prime}$ are similar with a ratio of 3, and $Z^{\prime} Z_{\alpha} = Y^{\prime} Y_{\alpha} = X X_{\alpha}$, we have $Z Z_{\alpha} = X X_{\alpha} + 3(Y Y_{\alpha} - X X_{\alpha}) = 3 Y Y_{\alpha} - 2 X X_{\alpha}$. (The difference $Y Y_{\alpha} - X X_{\alpha}$ can be either positive, as shown in the figure, or negative.)

We obtain that the distances from $Z$ to the faces $ABC, ABD, ACD, BCD$ are respectively $3 \cdot 15 - 2 \cdot 14 = 17, 3 \cdot 13 - 2 \cdot 11 = 17, 3 \cdot 25 - 2 \cdot 29 = 17, 3 \cdot 11 - 2 \cdot 8 = 17$. Moreover, it is clear that the point $Z$ is on the same side of each face as points $X$ and $Y$ (otherwise, the formula would give us a negative distance), which means it also lies inside the tetrahedron.

A point inside the tetrahedron, the distances from which to its faces are equal, is unique - it is the center of the inscribed sphere. The radius of the sphere is equal to the distance from the center to the faces, which is 17.

![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-21.jpg?height=363&width=618&top_left_y=85&top_left_x=141)

(a) tetrahedron

![](https://cdn.mathpix.com/cropped/2024_05_06_85a336fa11c4e8eb26a1g-21.jpg?height=244&width=460&top_left_y=201&top_left_x=834)

(b) section $X Y Y_{\alpha} X_{\alpha}$

Fig. 9: to the solution of problem 11.8