What's the largest regular tetrahedron (having side length $x$) you can fit inside a sphere with a unit radius?
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$\begingroup$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:
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
$\endgroup$ 2 $\begingroup$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 $AB=\sqrt(8/3)$.
$\endgroup$ $\begingroup$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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