设正四面体S-ABC,高SH,其中H是底面三角形ABC的外（内、重、垂）心,连结AH,在平面SAH上作SA垂直平分线,交SH于O,则O是内切（外接）球心,
设棱长为a,AH=a（√3/2）*（2/3）=a√3/3,
SH=√[a^2-(a√3/3)^2=a√6/3,
△SMO∽△SHA,设外接球半径=R,内切球半径=r,
SM*SA=SO*SH,a^2/2=R*a√6/3,
R=a√6/4,
r=SH-SO=a√6/3-a√6/4=a√6/12.