How do I calculate the height of a regular tetrahedron having side length $1$ ?
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?
Geometry – How to Calculate the Height of a Tetrahedron
geometrypolyhedra
Related Solutions
We are clearly looking for the regular tetrahedron inscribed in a sphere of radius 1 (i.e. with all its vertices lying on the sphere). The neat trick with regular tetrahedra is to inscribe them in a cube.
Wikipedia has a picture of the two regular tetrahedra you can find in a cube: http://en.wikipedia.org/wiki/File:CubeAndStel.gif
The cube inscribed in a unit sphere has side length $\frac{2}{\sqrt 3}$, so the regular tetrahedron has side length $x = \sqrt{2} \frac{2}{\sqrt 3} = \sqrt{\frac{8}{3}}$.
Here is another example of this idea: Height of a tetrahedron
One method is to divide the tetrahedron into 8 identical sections or two sets of 4 identical sections and then to find the smallest bounding cube as the sections are randomly rotated. For 8 cubes, I haven't been able to beat 1/2 yet.
Covering an edge-2 tetrahedron with 4 unit cubes is easy (link to code). With some twisting and optimization, the tetrahedron can be covered with 4 cubes with edge length 0.969975. This can likely be improved by allowing the cubes to interact.
Note that the question for how best to cover a triangle with two squares wasn't solved correctly until 2009. Instead of jumping to 8 cubes, it might be best to settle 2 cubes, 3 cubes, and 4 cubes first.
For a tetrahedron with edge length 2 ...
1 cube of edge length 1.41421356237309504
2 cubes of edge length 1.41421356237309504 (??)
3 cubes of edge length 1.3233 (below)
4 cubes of edge length 0.969975 (above)
5 cubes of edge length .848528 (below)
Here's a picture for 3 cubes. The tetrahedron has a base triangle ABC. One cube covers points A, D, mid-AB, mid-AC, and the center of the base. But there is likely improvement by shifting the points mid-AB, mid-AC.
Five cubes seems to have a simple solution.
Best Answer
The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.
Here's a view of the geometry:
and here's a view of the bottom face:
In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.
Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.
Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):
$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$
Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated here.
From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies
$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$
Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned here.