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.
I'll use the same lettering as Brian Smith but my method will differ.
O is the center of the sphere. AB=AC=BC=BD=CD=3, AD=x, M is the midpoint of BC.
Call P the center of equilateral triangle ABC. Then OB=2, PB=sqrt(3) so OP=1. PM=sqrt(3)/2 and OM=sqrt(7)/2.
Now take the plane section through the tetrahedron through triangle ADM, which by symmetry also contains O.
This triangle has sides AM=DM=3sqrt(3)/2, AD=x, OA=OD=2, OM=sqrt(7)/2
Law of Cosines on triangle AOM
cos(<AOM)=cos(<BOM)=(2^2+(sqrt(7)/2)^2-(3sqrt(3)/2)^2)/(2*2*sqrt(7)/2)=-sqrt(7)/14
A bit of identities gets
cos(<AOB)=-13/14
Law of Cosines on triangle AOD
x^2 = 2^2+2^2-2*2*2*cos(<AOB)
x^2 = 108/7
x=sqrt(108/7)= 6sqrt(21)/7
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Posted by Jer
on 2024-04-03 08:56:05 |
Exact solution
