What's the largest regular tetrahedron (having side length $x$) you can fit inside a sphere with a unit radius?

  • $\begingroup$ It's an extension of a question I saw in a textbook, not class work. $\endgroup$ – Patrick Beardmore Nov 28 '10 at 22:22
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    $\begingroup$ I'm not sure I understand the question. There is only one regular tetrahedron that can be inscribed in a sphere of a fixed radius. So... the largest is the one and only tetrahedron. $\endgroup$ – Ryan Budney Nov 28 '10 at 22:34
  • $\begingroup$ @Patrick Beardmore: do you want a proof of the formula stated in the Wikipedia? $\endgroup$ – Américo Tavares Nov 28 '10 at 22:45
  • $\begingroup$ Probably, some form of the "Purkiss principle" (thanks @Bill Dubuque!) might help here; since four points determine a sphere and a tetrahedron, the optimization of an appropriate objective function should give the result that the Platonic tetrahedron is optimal. $\endgroup$ – J. M. is a poor mathematician Nov 28 '10 at 23:01
  • $\begingroup$ I apologise. I had not seen the formula on Wikipedia. It is a proof of this that I'm looking for, and preferably multiple ways of looking at the problem. $\endgroup$ – Patrick Beardmore Nov 28 '10 at 23:33

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


You are looking at the regular tetrahedron inscribed in a sphere of radius 1. Denote the center of the sphere by O, and the vertices by A, B, C and D.

Fact: In the regular tetrahedron, the altitude from A is cut by O in 3:1 ratio (Note: In an equilateral triangle the analogous ratio is 2:1).

Proof: The four vectors from O to the vertices sum to zero (the sum vector is invariant under rotations preserving the tetrahedron, so must be zero). Pick a vertex A. The projections of the other three vectors OB, OC and OD onto the line OA are all equal and, as their sum must cancel OA, each of these projections is equal to 1/3 of OA.

Denote the base of the altitude by H. Then we just showed OH=1/3, AH=4/3. Now by Pythagoras theorem, $HB^2=BO^2-OH^2=1^2-(1/3)^2=8/9$. Finally, by Pythagoras again, $AB^2=AH^2+HB^2=(4/3)^2+8/9=24/9$, so $AB=\sqrt(8/3)$.


Notice that the radius $R$ of the circumsphere (i.e. spherical surface passing through all four vertices) of a regular tetrahedron, having an edge length $x$, is given as $$R=\frac{x}{2}\sqrt{\frac{3}{2}}$$
(Note: here is derivation of circumradius $R$)

Now, substituting radius of sphere $R=1$ in the above expression, we get $$1=\frac{x}{2}\sqrt{\frac{3}{2}}$$ $$\color{blue}{x=2\sqrt{\frac{2}{3}}\approx 1.632993162}$$


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