Referring to image:
Or alternatively you can view image at:
http://img182.echo.cx/img182/5145/trtrahedron8pt.jpg
In here I'll try to derive the circu-radius and In-radius for Tetrahedron with following premises (this I don't think we need to prove):
1. The altitude form the top vertex intersect the bottom face (Equilateral Triangle) at its centroid.
2. The circum-center lies on the altitude defined above.
Let side of a face of TetraHedron ABCD = 1
BL (altitude to Triangle BCD) = sqrt(3)/2
BO = circumradius of Triangle BCD = 2/3 * BL = 1/sqrt(3)
In right triangle ABO (right angled at O)
AB = 1
AO^2 = AB^2-BO^2
AO = sqrt(2/3)
Now we need to find point P on line AO such that AP = BP (then P will be the cicumcenter of the tetrahedron)
let OP = x
AP^2 = (AO- BO)^2
BP^2 = OP^2 + BO^2
and AP^2 = BP^2
(sqrt(2/3) - x)^2 = x^2 + 1/3
2/3 - 2*x* sqrt(2/3) = 1/3
x= 1/6*sqrt(3/2) = 1/6*sqrt(6/4)=sqrt(6)/12
AP = AO - x
AO = sqrt(2/3) = sqrt(6/9)=sqrt(6)/3
AP = sqrt(6)/4
hence cicumradius = sqrt(6)/4
OP will become inradius as this is perpandicular distance of a face from centerof tetrahedron, radius(incirlce) = sqrt(6)/12
Small sphere diameter = circumradius-inradius (this will happen when you drop a perpandicular from center to either face) = sqrt(6)/6
radius (small circale) = sqrt(6)/12
probability = (4* radius (small circale)^3 + radius(incircle)^3)/ circumradius ^3 = 5/27
(Taking volumes)
Linear Mode
