Create a tetrahedron starting with a square ABCD (clockwise), add 2 points E and F placing them halfway between AB and AD respectively.
Fold along lines EF, EC and FC.
Then, form the tetrahedron by keeping triangle EFC as the base, and folding triangles AEF, BEC, and DFC up so that points A, B, and D all meet.
For AB=a find the radius of the largest sphere contained by this tetrahedron
If the right triangle EAF is made to be the base, the edge from C at the vertex to the coincident A,B,D at the right angle on the base is vertical. Taking a vertical plane including this vertical edge and perpendicular to the hypotenuse of the base and therefore also perpendicular to the only slanting face of the tetrahedron (the others are the base and two vertical faces), and taking a crosssection of the sphere, we get the following facts:
The sphere of radius r is r units from either of the vertical faces and therefore r*sqrt(2) units from the vertical edge. Its distance from a point directly above the hypotenuse on the base on the crosssection plane is r/tan(D/2) where D is the dihedral angle at the hypotenuse of the base. Together r*sqrt(2)+r/tan(D/2) equals a/sqrt(2), which is the distance from the vertical edge to the hypotenuse of the base, along the base on the crosssection plane.
Based on the triangles, constructing a sphere about one end of the hypotenuse of the base, and taking a spherical triangle from where the faces of the tetrahedron intersect the sphere, and calling the side of the triangle opposite the angle that represents the dihedral angle a, cos(a)=1/sqrt(5), cos(b)=sqrt(2)/(2*sqrt(5)), cos(c)= 1/sqrt(2).
Also sin(b)=(2*sqrt(2)1/sqrt(2))/sqrt(5) and sin(c)=1/sqrt(2).
Given that D = A in the conventional labelling of triangles,
cos a = cos b cos c + sin a sin b cos D
Solving, cos D comes out to be 1/3.
Using tan(D/2) = sqrt((1cos D)/(1+cos D)), this comes out to 1/sqrt(2).
So r sqrt(2) + r sqrt(2) = a/sqrt(2)
4 r = a
r = a/4
unless I've made an algebraic error somewhere.

Posted by Charlie
on 20050419 20:30:16 