Problem 4. Given a triangular pyramid $A B C D$, where angles $B D C, C D A$, and $A D B$ are right angles and $D A=a$, $D B=b, D C=c$. Point $M$, belonging to the face $A B C$, is equidistant from the planes of all other faces of the pyramid. Calculate the length of the segment $D M$.
Solution. Let, for definiteness, $a \leqslant b \leqslant c$. We extend the given pyramid $A B C D$ to a cube with bases $A^{\prime} D B^{\prime} K^{\prime}$ and $A_{1} C B_{1} K_{1}$ (Fig. 4). Each point on the diagonal $D K_{1}$ of this cube is equidistant from the planes $C D B, C D A$, and $A D B$. Therefore, the point $M$ is the intersection of the segment $D K_{1}$ with the plane $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-163.jpg?height=642&width=688&top_left_y=296&top_left_x=341)

Fig. 3

Fig. 4

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-163.jpg?height=697&width=562&top_left_y=248&top_left_x=1158)

Let the lines $D K^{\prime}$ and $A B$ intersect at point $K$. Then $M$ is the intersection of the lines $C K$ and $D K_{1}$.

Triangles $M D K$ and $M K_{1} C$ are similar. Clearly, $C K_{1}=c \sqrt{2}, D K_{1}=c \sqrt{3} ; \quad M K_{1}=D K_{1}-D M=c \sqrt{3}-$ $-D M$. From the similarity of triangles $M D K$ and $M K_{1} C$ we get:

$$
\frac{D K_{1}-D M}{D M}=\frac{C K_{1}}{D K} \text { or } \frac{c \sqrt{3}}{D M}=\frac{c \sqrt{2}}{D K}+1 .
$$

Now it is clear that to solve the problem, we need to first find the length of the angle bisector $D K$ of the right triangle $A D B$.

The angle bisector $D K$ divides the segment $A B$ into parts proportional to the segments $D B$ and $D A$. Therefore,

$$
\overline{D K}=\frac{a}{a+b} \overline{D B}+\frac{b}{a+b} \overline{D A}
$$

Vectors $\overline{D A}$ and $\overline{D B}$ are perpendicular. Therefore,

$$
\begin{aligned}
\overline{D K}^{2}= & \frac{a^{2}}{(a+b)^{2}}{\overline{D B^{2}}}^{2}+\frac{b^{2}}{(a+b)^{2}} \overline{D A^{2}}=\frac{a^{2} b^{2}}{(a+b)^{2}}+ \\
& +\frac{b^{2} a^{2}}{(a+b)^{2}}=\frac{2 a^{2} b^{2}}{(a+b)^{2}}, D K=\frac{a b \sqrt{2}}{a+b}
\end{aligned}
$$

From equalities (1) and (2) we get:

$$
\begin{gathered}
\frac{c \sqrt{3}}{D M}=\frac{(a+b) c \sqrt{2}}{a b \sqrt{2}}+1=\frac{a c+b c+a b}{a b} \\
\text { and } D M=\frac{a b c \sqrt{3}}{a c+b c+a b}
\end{gathered}
$$