# Volume of Sphere Inscribed within a Tetrahedron

• Jul 15th 2009, 05:37 PM
nickerus
Volume of Sphere Inscribed within a Tetrahedron
So we have a tetrahedral pyramid with side length a. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid?

I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that all the faces are equilateral triangles.

Start by solving for the height of each triangular face, with side length a.
$a^2+(\frac{a}{2})^2=k$
where k is the height of one of our face triangles.
Solving, $k=\frac{a\sqrt{3}}{2}$
Now, we can use k to solve for the height of the pyramid, h; imagine a new triangle cutting the pyramid in half.
$k^2=(\frac{k}{2})^2+h^2$
Solving, $h=\frac{k\sqrt{3}}{2}=\frac{3a}{4}$
Using the relationship between a 2-d triangle's height and an inscribed circle's diameter, which is
$height=\frac{3}{2}*diameter$,
$diameter=\frac{2}{3}*height$,
we can find the diameter, d, and radius, r, of the inscribed sphere.
$r=\frac{d}{2}=\frac{h}{3}=\frac{3a}{4}*\frac{1}{3} =\frac{a}{4}$
Plug the radius, r, into the volume of a sphere, V, to get the solution.
$V=\frac{4}{3} \pi r^3=\frac{4}{3}\pi (\frac{a}{4})^3=\frac{4\pi}{3}*\frac{a^3}{64}=\fra c{\pi a^3}{48}$
...and that's the solution I've come up with. And I have no idea how to even check to see if it's correct. I'll try to get a picture up so the solution is more visible.
• Jul 16th 2009, 01:46 AM
earboth
Quote:

Originally Posted by nickerus
So we have a tetrahedral pyramid with side length a. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid?

I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that all the faces are equilateral triangles.

Start by solving for the height of each triangular face, with side length a.
$a^2+(\frac{a}{2})^2=k$
where k is the height of one of our face triangles.
Solving, $k=\frac{a\sqrt{3}}{2}$
Now, we can use k to solve for the height of the pyramid, h; imagine a new triangle cutting the pyramid in half.
$k^2=(\frac{k}{2})^2+h^2$ ..... <<<<< this can't be correct. See attachment.

...

...

In my opinion the equation should read:

$k^2=\left(\dfrac13 k\right)^2 + h^2$

or

$a^2=\left(\dfrac23 k\right)^2 + h^2$
• Jul 16th 2009, 04:57 PM
nickerus
Yeah I think that's right; I don't know why I went with k/2 hehe. Then, the height of the pyramid would be
$h=\frac{2k\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}*\frac{a \sqrt{3}}{2}=\frac{a\sqrt{6}}{3}$,
and the radius of the sphere would be
$r=\frac{h}{3}=\frac{1}{3}*\frac{a\sqrt{6}}{3}=\fra c{a\sqrt{6}}{9}$.
But then the volume works out to a pretty gnarly fraction,
$V=\frac{8\pi a^3 \sqrt{6}}{729}$
On wikipedia, it lists the radius of the insphere of a tetrahedron as $\frac{a}{\sqrt{24}}$, so how did they get that..? I could always just be doing my algebra wrong.
• Jul 17th 2009, 04:53 AM
alunw
I've tried working things like this out before now and its usually not that easy.
Here's how I'd go about it for a tetrahedron.
If you take an ordinary cube, and pick two vertices diagonally opposite each other on one face, and the other two vertices diagonally opposite each other on the opposite face you get a regular tetrahedron.
This shows that the shape with vertices at say (1,1,1),(1,-1,-1),(-1,-1,1),(-1,1,-1) is a regular tetrahedron. The side of the cube is 2, and the side of the tetrahedron is 2*sqrt(2) = sqrt(8). The point of tangency of the sphere to one of the faces is easy to calculate
as well. For example one of them is 1/3 of the way along the line from (1,0,0) (the midpoint of a side ) to (-1,1,-1) (one of the two opposite vertices), i.e at (1/3,1/3,-1/3) since the sphere is surely tangent at the centre of the face. Now we certainly hope the centre of the inscribed sphere is at (0,0,0) and by considering all the other possible tetrahedra we could inscribe in the cube we see this must be so.

So the radius of the sphere is sqrt(1/3) and sqrt(1/3)/sqrt(8) is sqrt(1/24).