The length of five edges of a tetrahedron inscribed in a sphere of radius 2 is 3. Find the length of the sixth edge of the tetrahedron.
Let ABCD be the tetrahedron with AD being the one side that is not equal to 2. Then AB=AC=BC=BD=CD=2.
Let O be the center of the sphere.
Let M be the midpoint of BC. Then AM=MD=sqrt(3).
Now thanks to symmetry A, D, M, and O are all coplanar. Then O is the circumcenter of triangle ADM, and the circumradius is 3.
A formula for circumradius for a triangle with sides x, y, z is:
R = x*y*z/sqrt[(x+y+z)*(x+y-z)*(x-y+z)*(-x+y+z)]
Then for this problem, x=y=sqrt(3), R=3 and z is the unknown length AD.
3 = 3z/sqrt[(2sqrt(3)+z)*(2sqrt(3)-z)*z^2]
Rearrange and do a bit of cancelling:
sqrt[(2sqrt(3)+z)*(2sqrt(3)-z)] = 1
Then square each side and simplify to get z^2 = 11, or z=sqrt(11).
Solution
