What's the largest regular tetrahedron (having side length $x$) you can fit inside a sphere with a unit radius?
[Math] Tetrahedron inside a sphere
geometrypolyhedra
Related Solutions
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.
The center of the tetrahedron divides each of the four heights (or medians) in the ratio $1:3$ (in an equilateral triangle the corresponding ratio is $1:2$). The smaller part is also the radius of the inscribed sphere. Therefore the radius of this sphere is one quarter of the height or $9$ in your case.
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Best Answer
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