Radius of inscribed sphere within regular tetrahedron A question says:

A sphere is inscribed in a regular tetrahedron. If the length of an altitude of the tetrahedron is 36, what is the length of a radius of the sphere?

I'm not sure where to start.
This is what I think so far:


*

*I think that the sphere touches the "center" of each of the tetrahedron's sides.

*Halfway down one of the tetrahedron's sides, where it meets the "altitude line" perpendicularly, is the radius of the sphere.


Apparently, the answer's 9.
 A: 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.
A: An intuitive way of thinking about this is to realize that the centroid of the tetrahedron (and thus the sphere) is the average of the positions of the four corners. If three of the four corners are on the ground (I.e., altitude of 0), and the fourth corner has an altitude of 36, then it's pretty simple to figure out how high off the ground the centroid is. You want the radius of the sphere inscribed in the tetrahedron? That's the distance from the centroid to the ground. Want the radius of the sphere that inscribes the tetrahedron? That's the distance from the centroid to the top corner.
A: A tetrahedron with face area $F$ and altitude (to that face) $a$ has volume $\frac{1}{3}a F$.
Now, consider a sub-tetrahedron determined by a given face, with fourth vertex at the center of the inscribed sphere. Because the sphere is tangent to the face, the altitude to that face is a radius, say, of length $r$. If the big tetrahedron is regular (with face area $F$ and altitude $a$), then the four sub-tetrahedra are congruent, with volume $\frac{1}{3}r F$.
Together, the sub-tetrahedra fill the big tetrahedron, so the four sub-volumes add up to the big volume:
$$4 \cdot \frac{1}{3} r F = \frac{1}{3} a F$$
so that $4 r = a$. You're given that $a=36$; consequently, $r = 36/4 = 9$.

Note. Regular or not, the basic argument shows that the volume ($V$) of a tetrahedron is determined by its inradius ($r$) and its total surface area ($S$):
$$V = \frac{1}{3} r S$$
This is analogous to the fact that the area ($A$) of a triangle is given by the inradius ($r$) and its perimeter ($p$):
$$A = \frac{1}{2}r p$$
The pattern continues into $d$-dimensional space, with the multiplied constant being $\frac{1}{d}$.
A: To derive the results from first principles, I would name the side length of the tetrahedron $s$ and work in multiples of $s$ until I had expressions for both the altitude and the radius of the inscribed sphere.
First step is to find some lengths in each of the equilateral triangles that form the tetrahedron's sides -- in particular the distance between the midpoint and a corner, and the perpendicular distance from the midpoint to one of the edges. Both are easily found by dividing the triangle into appropriate 30-60-90 triangles.
In the second step we move to 3D. By symmetry, the center of the inscribed sphere must lie on all four altitudes (one from each corner), so we can find it exactly by intersecting two altitudes. So we work in the plane that contains two of the altitudes. This plane contains two corners A and B, and also the midpoints of the opposing sides, and also the midpoint of the edge between the two opposing sides. Thus, the cross section of the tetrahedron by this plane contains several lengths that we already know from the first steps. There are no nice well-known angles in the cross section, but the lengths are enough to apply of Pythagoras and similar triangles to find the two lengths you're ultimately interested in.
A: Another way of looking at this in a way very similar to the answer of IV is the following.
A suitable affine motion will place the tetrahedron in such a position that its vertices are at points $(0,0,0)$, $(a,a,0)$, $(a,0,a)$ and $(0,a,a)$. We further note that these points are also vertices of the cube $[0,a]^3$. Here the symmetries of the position dictate that the center of the inscribed sphere is at the center of the cube, i.e. at the point $(a/2,a/2,a/2)$.
Let us look at the altitude starting from the origin. The other three vertices are on the plane $x+y+z=2a$, so the altitude is parallel to the vector $\vec{n}=(1,1,1)$. The centroid is on the plane $x+y+z=3a/2$ also normal to $\vec{n}$. Therefore the centroid splits the altitude in $3:1$ ratio. The claim follows.
