Prove. The sum of the distances from a random point in a regular tetrahedron to each of its walls = *h*

Question

We have a random point A in a regular tetrahedron. Prove that the sum of the distances from A to each wall is equal to the height of the tetrahedron.

Answer 1

This is an extension of Viviani's Theorem:

Consider the $4$ tetrahedrons formed by drawing the segments from $A$ to the vertices of the original tetrahedron. Let the individual distances to the faces be $h_1, h_2, h_3, h_4$.

For each face of the tetrahedron, the volume of the sub-tetrahedron with vertex at $A$ is $h_iF/3$, where $F$ is the area of a face. Because the sum of the volumes is the total volume,

$$\frac{1}{3} \cdot F \sum h_i = V.$$ Now consider the height $h$ of the tetrahedron. By the volume formula again,

$$V = \frac{1}{3}hF$$

Thus, $\sum h_i = h$.