Problem:
Let ABCD be a tetrahedron such that AB=CD=41,AC=BD=80, and BC=AD=89. There exists a point I inside the tetrahedron such that the distances from I to each of the faces of the tetrahedron are all equal. This distance can be written in the form pmn, where m,n, and p are positive integers, m and p are relatively prime, and n is not divisible by the square of any prime. Find m+n+p.
Solution:
The required distance is the radius, r, of the insphere of the tetrahedron. Note that △ABC,△BAD,△CDA, and △DCB are congruent, and thus they have the same area. Let this shared area be K. This implies that the distances from any one of the vertices to the base containing the other three vertices must all be equal. Let this value be h. The volume of tetrahedron ABCD is then 31hK, and it is also equal to the sum of the volumes of tetrahedra ABCI,ABDI,ACDI, and BCDI. These tetrahedra have a base of area K and a height r, so the volume of tetrahedron ABCD is also 4⋅31rK. Therefore 31hK=4⋅31rK, so r=4h.
Let u,v, and w be positive real numbers that satisfy the equations
AB2=41AC2=80BC2=89=u2+v2,=v2+w2, and=u2+w2.
Placing points A,B, and C in 3-dimensional coordinate space at A(0,0,0),B(u,v,0),C(0,v,w) results in segments AB,AC, and BC having the required lengths. In addition, if point D is chosen to be D(u,0,w), then
and ABCD is the required tetrahedron. The system above has a unique solution u=5,v=4, and w=8, and thus the vertices of the tetrahedron are A(0,0,0),B(5,4,0),C(0,4,8), and D(5,0,8). Vertices B,C, and D lie in the plane with equation
5x+4y+8z=2.
The distance from a point (x,y,z) to the plane with equation ax+by+cz+d=0 is given by
a2+b2+c2∣ax+by+cz+d∣
The altitude of the tetrahedron is the distance from A to the plane containing B,C, and D, which is
h=521+421+8212=638021.
The required inradius of the tetrahedron is 4h=632021. The requested sum is 20+21+63=104.
OR
As in the first solution, three of the edges of the tetrahedron can be represented by the vectors ⟨5,4,0⟩, ⟨0,4,8⟩, and ⟨5,0,8⟩. The area of one of the faces of the tetrahedron can be calculated using Heron's Formula or by calculating one half the length of the cross product of two of these vectors:
21⋅∣⟨5,4,0⟩×⟨0,4,8⟩∣=621.
The given tetrahedron fits in a rectangular box with dimensions 5×4×8, and the volume of the tetrahedron is the volume of this box minus the volumes of four congruent corners. The volume of each corner is 31⋅21⋅5⋅4⋅8, so the volume of the tetrahedron is
5⋅4⋅8−4⋅65⋅4⋅8=3160.
The volume of the tetrahedron can also be calculated as one sixth the triple product of the vectors:
61∣∣∣∣∣∣∣505440088∣∣∣∣∣∣∣=3160.
As in the first solution, the required distance, r, is one quarter of the altitude of the tetrahedron, so the volume of the tetrahedron is
31⋅4r⋅621=3160,
from which r=632021, as in the first solution.
Note: The volume of a tetrahedron with known edge lengths can also be calculated using Tartaglia's Formula. Also, the square of the volume of an isosceles tetrahedron with side lengths a,b, and c can be written as 721(a2+b2−c2)(b2+c2−a2)(c2+a2−b2).