Geometry of the regular tetrahedron
A regular tetrahedron is a shape consisting of four equilateral triangles (triangles with all their sides equal). A tetrahedron has four vertices, six edges, and four faces. Let the length of the edges of the regular tetrahedron be one. Then the length of the altitude, exemplified by OD in the diagram (the perpendicular distance from one face of the tetrahedron to the other vertex) of the tetrahedron is \(\sqrt{2/3}\). The point where the altitude of the vertex meets the face, O, is \(1/\sqrt{3}\) from each of the three vertices A, B, and C of the face, and \(1/2\sqrt{3}\) from the nearest point of the edges M.
Proof
The altitude of the faces
Here is one of the faces of the tetrahedron. The point where the altitude meets the face, labelled O, must, by symmetry, be the same distance from each of the vertices A, B, and C.
The red lines in the figure are also called the altitudes of the triangle. The midpoint of AB is M. The points A, C, and M form a right angled triangle. AM is the midpoint of AB, so its length is 1/2. The length of the altitude AM is, by Pythagoras's theorem
\[ \sqrt{1-\left(\frac12\right)^2}=\frac{\sqrt3}2. \]The distance to O
The vertex of the tetrahedron opposite this face, D, is equidistant from A, B, and C, so the point O where the altitude meets the triangle ABC must be on the meeting point of the face's altitudes. The altitudes meet at the centroid of the triangle, so the length from the midpoint M to the centroid O is a third of the length of the altitude, or \(1/2\sqrt3\).
Another way to calculate OM is that the total area of ABC is \(1\times\sqrt3/2\). By symmetry each of the right angled triangles demarcated by the red lines is similar. Thus the area of ABC is six times the area of one triangle, so the altitude of the right angled triangle \(6\times 1/2\times OM=\sqrt3/2\), and so \(OM=\sqrt{\frac13}/2\). Then the distance from the three vertices to O is \(\sqrt{1/4+1/12}=\sqrt{1/3}\).
Yet another way to see this is that the length OA = OC, thus
\[ OA^2=AM^2+OM^2=(CM-OM)^2=CM^2-2CMOM+OM^2 \]and so \[ OM=\frac{CM^2-AM^2}{2CM}=\frac{1}{2\sqrt{3}} \]
The height
The altitude of the tetrahedron OD, the edge of the tetrahedron CD, and the line OC form another right-angled triangle, and the length of the altitude OD is, by Pythagoras's theorem, then equal to \(\sqrt{1-1/3}=\sqrt{2/3}\).