Sphere inscribed in a regular tetrahedron

• Mar 9th 2010, 04:41 PM
spred
Sphere inscribed in a regular tetrahedron
Find the radius of a sphere inscribed in a regular tetrahedron which has a height of 8.

Thank You
• Mar 10th 2010, 05:22 AM
Hello spred
Quote:

Originally Posted by spred
Find the radius of a sphere inscribed in a regular tetrahedron which has a height of 8.

Thank You

The attached diagram shows a vertical cross-section through the tetrahedron, along one of the medians, $\displaystyle AM$, of the base.

$\displaystyle V$ is the vertex of the tetrahedron, and $\displaystyle O$ the centre of the inscribed sphere.

$\displaystyle G$ is the centroid of the base; $\displaystyle H$ is the centroid of one of the sloping faces. By symmetry, the sphere touches the base and this face at $\displaystyle G$ and $\displaystyle H$.

The height of the tetrahedron is $\displaystyle h$. So $\displaystyle VG = AH = h$. (In your question $\displaystyle h = 8$.)

Also $\displaystyle GM = HM = \tfrac13AM$, using the well-known property of the centroid of a triangle. So if $\displaystyle \angle HAM = \theta$,
$\displaystyle \sin \theta = \frac{HM}{AM}= \frac13$

$\displaystyle \Rightarrow \tan\theta = \frac{1}{\sqrt8}$
...(1)

$\displaystyle \Rightarrow GM =HM = h\tan\theta$

$\displaystyle \Rightarrow AG = 2GM=2h\tan\theta$

$\displaystyle \Rightarrow r = OG = AG \tan\theta$
$\displaystyle =2h\tan^2\theta$

$\displaystyle =\tfrac14h$, from
(1)
So the radius of the inscribed sphere is one-quarter of the height of the regular tetrahedron. In your question, then, the radius is $\displaystyle 2$ units.