A tetrahedron with five edges of unit length is inscribed in a unit sphere. How long is the sixth edge?
Each of the five given edges subtends an angle at the center of the sphere, and therefore an arc, of 2*arcsin(1/2) = 60°.
Each of the two spherical triangles formed by the five given edges (two faces of the tetrahedron) is equilateral with 60° edges. A complete 360° great circle can be formed consisting of the heights of these two equilateral triangles from their common base plus the arc that has the unknown edge as a chord. So first we need to find the altitude of each of the spherical triangles and subtract twice that value from 360°. Then we can find the chord subtended by that arc.
Either of the two equilateral spherical triangles can be divided by an altitude into two right triangles. One leg is the unknown altitude and the other leg is 30°. The hypotenuse is 60°. We use the spherical law of cosines:
cos(60°) = cos(30°)*cos(x) + sin(30°)*sin(x)*cos(90°)
cos(x) = cos(60°)/cos(30°) ~= .577350269189627 = sqrt(1/3)
While the angular size of this arc (about 54.74°) does not directly interest us, we want to know the length of its subtended chord, L.
**** no, no, no; we need to go further; see my corrective post ****
The rest of this post is wrong!
L = 2*sin(x/2)
This is approximately .919401686761962.
We can state this explicitly, by substituting sqrt(1/3) as cos(theta) into the sine of half angle formula, as
2 * sqrt((1-sqrt(1/3))/2)
Edited on April 21, 2018, 5:21 pm
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Posted by Charlie
on 2018-04-20 13:18:39 |
solution
