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.
Project the edges of the tetrahedron onto the circumscribing sphere. A chord of length 3 projects to an arc, a, which can be found as
sin(a/2) = 1.5/2
where 1.5 is half the chord length and 2 is the sphere's radius. So a = 2*arcsin(.75) =~ 97.1807557814583 °.
The projected angle is then found on the spherical triangle whose sides are all equal to a.
cos(a) = cos(a)^2 + sin(a)^2 * cos(A)
-1/8 = 1/64 + (63/64) * cos(A)
cos(A) = (-1/8 - 1/64) / (63/64)
A = 98.2132107017382
The arc projected from the nonequal edge (chord) from the tetrahedron (call it a') is found from
cos(a') = cos(a)^2 + sin(a)^2 * cos(360 - 2*A)
as the projection of the vertex opposite the unequal side is a meeting of two of the equal corners and one of the unequal corner.
cos(a') = 1/64 + (3*sqrt(7)/8) * cos(360 - 2*A)
a' =~ 158.213210701738
The edge (chord) in question is then
2 * sin(158.213210701738/2) * 2
=~ 3.92792202424786
|
|
Posted by Charlie
on 2024-04-02 14:54:26 |
what I get
