设正四面体棱长为a,顶点为A,高为AM,球心为O.则有AM^2=[(√3a)/2]^2-[(√3a)/6]^2 得AM=AO+OM=R+OM=(2a√6)/6①有OM/R=1/3②由得①②a=4R/(a√6)又因为可求底面S=[（√3)/4]*a^2 v=(1/3)*S底面*AM=(√2)/12a^3∴所求...