A Tetrahedron is a "Platonic Solid", or Regular Polyhedron. A regular polyhedron is convex, and its faces are all regular polygons. There are only five such solids, and the tetrahedron is the smallest of them. The others are the cube, the octahedron (with 8 equilateral triangle faces, made by gluing together the bases of two square pyramids with equal edge lengths), the icosahedron (with 20 equilateral triangular faces), and the dodecahedron (with 12 pentagonal faces).
A math puzzle site called "The Problem Site, The Maine Page" posed the following problem:
A sphere of radius X is inscribed in a regular tetrahedron of arbitrary side length, and a sphere of radius Y is circumscribed about the same tetrahedron. What is the numerical value of the ratio Y:X? Express your answer in the simplest possible form.
The crux of this problem is to find that magical point equidistant from each of the faces (the incenter) and the magical point equidistant from each of the vertices (the circumcenter). The highly symmetrical nature of the tetrahedron makes it "obvious" that the incenter and circumcenter are the same point, and in fact the centroid is also coincident with these.
A cone or pyramid (any solid figure whose cross-section area varies as the square of the distance from its apex) has a volume equal to 1/3 times the height (measured perpendicular to the base) times the area of the base. This is found by letting the area equal k(h-x)², where k is a constant, and h the height of the figure, and then integrating.
k(h-x)² dx = (1/3)(kh²)(h) = (1/3)Ah, where A is the area of the base.
The centroid of any such figure is the ratio of the following integrals:
xk(h-x)² dx divide by
The top integral is (1/12)(kh²)(h²), and the bottom one, as we have seen, is (1/3)(kh²)(h). So their ratio is h/4. That is, the centroid of a tetrahedron (or any conical or pyramidal shape, for that matter) is a quarter of the way from base to apex.
Thus the ratio of the circumscribed sphere (the distance from the centroid to the apex) is three times the ratio of the inscribed sphere (the distance from the centroid to the base).
This completes the puzzle, but for the sake of a fuller understanding of the puzzle, I will put the tetrahedron on a system of coordinates with the centroid at the origin, and use a spreadsheet to calculate the distances between all the points, and between the points and the origin. This will convince me that the answer is correct.
The ratio of the circumscribed sphere to the inscribed sphere in a tetrahedron is 3:1, because the centroid (which is also the incenter and circumcenter) is one quarter of the way up the height, measured from the base to the apex.
The centroid is easy to calculate using the ratio of two definite integrals, the numerator having each infinitesimal slice multiplied by x, and the denominator adding up the slices alone. For any conical or pyramidal shape, the centroid (or more exactly its coordinate perpendicular to the base) is a quarter of the way up from the base along the perpendicular (to the base) line that goes through the apex.
First, I'd like to simplify the math by letting the edges of the tetrahedron measure one "unit" in length. From now on, all the lengths referenced here will be measured in "units", areas in "square units", etc.
I'll arbitrarily pick one face of the tetrahedron and call it the "base". The base will be parallel to the x-y plane, and the origin will be at its center. The base of the base (the line segment forming the base of that triangular face) will be parallel to and "below" (in the y-dimension) the x axis. We know already (or we can figure out, using the method shown above) that the centroid of a triangle is one third of the way from its base to its apex, and the height of the unit equilateral triangle is sqrt(3)/2, so the y-coordinate of the base is -sqrt(3)/6. The x-coordinates range from -1/2 to 1/2, so two of the coordinates in the x-y plane of the tetrahedron are
x-y coordinates of ends of the base of the base
(-1/2, -sqrt(3)/6) and
The third corner of the base of the tetrahedron is on the y-axis, and it's twice as far from the origin as the base-of-the-base, so its coordinates are
x-y coordinates of the apex of the base
Now I need to "sink" the base of the tetrahedron down a quarter of the height of the tetrahedron into the z dimension so that the origin in three dimensions will be the very center of the tetrahedron. For this, I need to know the height of a tetrahedron.
Consider the right triangle whose base runs from the center of the base of the tetrahedron to one of its edges and whose height is the height of the tetrahedron. Its length is sqrt(3)/6 The hypotenuse of this triangle runs along one of the faces of the tetrahedron, so its length is sqrt(3)/2. By the Pythagorean Theorem, we calculate the height of the tetrahedron to be sqrt(2/3).
A cute trick for using the Pythagorean Theorem quickly:
Express each of the sides of the right triangle as the square root of something. Then the numbers inside the square roots will add up nicely. Using the example, above, if you express the base as sqrt(3/36) and the hypotenuse as sqrt(3/4) = sqrt(27/36) then the height is the sqrt of the difference of these two numbers -- sqrt(27/36 - 3/36) = sqrt(24/36).
Draw the triangle and label its three sides: sqrt(3/36), sqrt(24/36), and sqrt(27/36)
Now we "sink" the base of the tetrahedron down in the z dimension a quarter of sqrt(2/3), and we "raise up" the apex of the tetrahedron by (3/4)sqrt(2/3), giving us the four coordinates of the tetrahedron:
x-y-z coordinates of the vertices of the tetrahedron
(-1/2, -sqrt(3)/6, -sqrt(2/3)/4)
(1/2, -sqrt(3)/6, -sqrt(2/3)/4)
(0, 2sqrt(3)/6, -sqrt(2/3)/4)
(0, 0, 3sqrt(2/3)/4)
Using a spreadsheet, I verified that the six distances between any pair of these four points is 1, and the distance from each of these four points to the origin is 3sqrt(2/3)/4.
Mathworld -- Platonic Solids, Tetrahedron
The Problem Site -- The Maine Page was the source of the problem posed and solved here.
Go back to Geometry - Solids
Centroid describes what it is (the balancing point) and how to calculate it using integrals.
Platonic and Archimedean Solids
Solid of Rotation describes how to calculate it using integrals, and better yet, how to calculate it using Pappus' Centroid Theorem. It also has a table of surface areas and volumes of a variety of simple solid shapes.
The webmaster and author of this Math Help site is Graeme McRae.