Volume of a tetrahedron What is the volume of a tetrahedron given the distance (x) from the center of the tetrahedron to one of the vertices? I can't figure out a short, nice method from getting the answer, so hints/general methods would be appreciated.
 A: For a Platonic solid with Schläfli symbol $\lbrace p , q \rbrace$ and circumradius $R$ (the distance from center to any vertex), the volume is
$$V = \frac{F p R^3}{3 \left( \tan(\frac{\pi}{q})\right)^3 \, \left(\tan(\frac{\theta}{2})\right)^2 \, \left(\tan(\frac{\pi}{p})\right)^2} \tag{1}\label{1}$$
where the number of faces is $F$,
$$F = \frac{4 q}{4 - (p - 2) (q - 2)} \tag{2}\label{2}$$
and the dihedral angle is $\theta$,
$$\sin(\frac{\theta}{2}) = \frac{ \cos(\frac{\pi}{q}) }{ \sin(\frac{\pi}{p}) } $$i.e.
$$\frac{\theta}{2} = \arcsin\left( \frac{\cos(\frac{\pi}{q})}{\sin(\frac{\pi}{p})} \right) \tag{3}\label{3}$$
If you want to re-derive these formulas, look at the Wikipedia Platonic solid article. First, solve the edge length $a$ from the equation for the circumradius $R$; the formula is valid for all Platonic solids. (For platonic solids, the circumsphere is concentric with the polyhedron itself: the centers are at the same point.) The article then shows how one can split the regular convex polyhedron to a set of equal pyramids, and using that, calculate the volume of the polyhedron. Substitute the variables, and you end up with $\eqref{1}$.
A: Let $ABCD$ be the vertices of the regular tetrahedron, $O$ its center and $H$ the center of face $ABC$. Let's denote by $r=OA=OB=OC=OD$ the tetrahedron circumradius, and by $a$ the length of its edges.
The four pyramids $OABC$, $OACD$, $OBCD$, $OABD$ are all equal among them and their volume is $1\over4$ the tetrahedron volume. But pyramids $OABC$ and $ABCD$ have the base $ABC$ in common, so their heights $OH$ and $DH$ must be proportional to their volumes: 
$$
OH={1\over4}DH={1\over4}(r+OH),
\quad\hbox{whence:}\quad
OH={1\over3}r
\quad\hbox{and}\quad
DH={4\over3}r.
$$
On the other hand, by applying Pythagoras' theorem to triangles $AOH$ and $ADH$ we have:
$$
AH^2=AD^2-DH^2=AO^2-OH^2,
$$
that is:
$$
a^2-{16\over9}r^2=r^2-{1\over9}r^2,
\quad\hbox{whence:}\quad
a^2={8\over3}r^2.
$$
Once you know the ratio $a/r$ it's easy to compute the area of $ABC$ and then the volume of tetrahedron $ABCD$ in terms of $r$:
$$
area_{ABC}={1\over2}a{\sqrt3\over2}a={\sqrt3\over4}a^2={2\over\sqrt3}r^2,
$$ 
$$
volume_{ABCD}={1\over3}area_{ABC}\cdot DH=
{1\over3}{2\over\sqrt3}r^2\cdot{4\over3}r=
{8\over9\sqrt3}r^3.
$$ 
