Consider a tetrahedron with two isosceles triangle faces with side lengths 510,510,10 and two isosceles triangle faces with side lengths 510,510,18. The four vertices of the tetrahedron lie on a sphere with center S, and the four faces of the tetrahedron are tangent to a sphere with center R. The distance RS can be written as nm, where m and n are relatively prime positive integers. Find m+n.
By Hints 9 and 10 dR,ABC=dR,BCD42+32∣3r∣=122+52∣12r−60∣0=min{zA,zB,zC,zD}≤r≤max{zA,zB,zC,zD}=12253r=16960−5r⇒53r=1360−5r39r=300−25r64r=300⇒r=1675
By Hints 7 and 11 S(0,0,325)R(0,0,1675)RS=325−1675=48400−225=48175=nmAns=m+n=175+48=223
