0

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.

J. Doe
  • 41

1 Answers1

0

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, $\sum h_i = h$.

  • You just assume that h1+h2+h3+h4=H and: HF=3V hF=3V It's obvious that h=H, but i need the reason why h1+h2+h3+h4=h – J. Doe Jun 04 '18 at 14:22
  • You don't seem to get the argument. Let me make this more clear for you, –  Jun 04 '18 at 14:26