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
(In reply to
solution by Charlie)
"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."
Actually, the distance from the vertical edge to the hypotenuse of the base, along the base, is a/(2 sqrt(2)). Thus, near the end, the equation should be:
So r sqrt(2) + r sqrt(2) = a/(2 sqrt(2))
8 r = a
r = a/8

Posted by Charlie
on 20050420 15:27:30 