A regular tetrahedron holds a sphere snugly within its four sides. A larger sphere surrounds the tetrahedron, just touching its four vertices. What is the ratio of radii of the two spheres?

According to the cosines law, an angle theta can be obtained using the tetrahedron side L and the height of the triangles H. Since a tetrahedron has te same sides, it is made of 4 triangles with same sides L, so their internal angle is 60 degrees and is equal to √/2, H becomes √3*L/2.
So theta becomes ArcCos[1/3].
We make a new rectang triangle with an hypotenusa equals to the already used height H (√3*L/2) and the angle theta on one vertex. With these two values, we obtain the bottom of the triangle instead of the full height of the tetrahedron. The bottom of the triangle is H/3.
Subtracting H-(H/3) gives 2*H/3.
Finally we make a rectang triangle where the hipotenusa is R2 (Radius of the external sphere), the bottom side 2*H/3 and the side left R1 (Radius of the internal sphere), and by angle relation, the angle between R2 and R1 is the already described theta (ArcCos[1/3]). So we obtain R1= √3*L/(3*Tan[ArcCos[1/3]]).
R2=R1/(Cos[ArcCos[1/3]])=3*R1=√3*L/(Tan[ArcCos[1/3]]).

THE RATIO OF RADII OF THE TWO SPHERES IS (R2/R1)=3

Note: I prefered to use the trigonometric function ArcCos[theta] in this problem, but it can be done using Pitagora's theorem.

Posted by Antonio on 2003-08-21 19:09:41