Problem:
A sphere is inscribed in the tetrahedron whose vertices are A=(6,0,0), B=(0,4,0),C=(0,0,2), and D=(0,0,0). The radius of the sphere is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let r be the radius of the inscribed sphere. Because ABCD can be dissected into four tetrahedra, all of which meet at the incenter, have a height of length r, and have a face of the large tetrahedron as a base, it follows that r times the surface area of ABCD equals three times the volume of ABCD. To find the area [ABC] of triangular face ABC, first calculate AB=52,BC=20, and CA=40. Then apply the Law of Cosines to find that cos∠CAB=9/130. It follows that sin∠CAB=7/130, so that [ABC]=21⋅AB⋅AC⋅sin∠CAB=14. The surface area of ABCD is
[ABC]+[ABD]+[ACD]+[BCD]=14+12+6+4=36
The volume of tetrahedron ABCD is 31⋅2⋅21⋅4⋅6=8. Thus r=24/36=2/3 and m+n=5.
OR
Because the sphere is tangent to the xy-plane, the yz-plane, and the xz-plane, its center is (r,r,r), where r is the radius of the sphere. An equation for the plane of triangle ABC is 2x+3y+6z=12, so the sphere is tangent to this plane at (r+2t,r+3t,r+6t), for some positive number t. Thus 2(r+2t)+3(r+3t)+6(r+6t)=12 and (2t)2+(3t)2+(6t)2=r2, from which follow 11r+49t=12 and 7t=r, respectively. Combine these equations to discover that r=2/3 and m+n=5.
OR
An equation of the plane of triangle ABC is 2x+3y+6z=12. The distance from the plane to (r,r,r) is
22+32+62∣2r+3r+6r−12∣
This leads to 7∣11r−12∣=r, which is satisfied by r=3 and r=2/3. Since (3,3,3) is outside the tetrahedron, r=2/3 and m+n=5.
Query: The sphere determined by r=3 is outside the tetrahedron and tangent to the planes containing its faces. Can you find the radii of the other three spheres with this property?