The points are arranged into two squares on parallel planes. One square is roatated 45 degrees
relative to the other. This shape is known as a square antiprism.
Let the verticies of one square have coordinates (a,a,b), (a,-a,b), (-a,a,b), (-a,-a,b), with
a>0, b>0.
The verticies of the other square would have coordinates of (c,0,-d), (0,c,-d), (-c,0,-d),
(0,-c,-d), with c>0, d>0.
Since all the points are on a sphere with radius 1, then two equations can be formed:
2a^2 + b^2 = 1
c^2 + d^2 = 1
The distance is maximized when the edges are equal, otherwise there would be some 'wiggle room'
to increase shorter edges.
2a = c*sqrt(2) = sqrt((c-a)^2 + a^2 + (b+d)^2)
Rewrite the compound equation as:
4a^2 = (c-a)^2 + a^2 + (b+d)^2
a*sqrt(2) = c
Substitute the expression for c into the other equations:
2a^2 + b^2 = 1
2a^2 + d^2 = 1
4a^2 = (a*sqrt(2)-a)^2 + a^2 + (b+d)^2
From the first two equations, b=d. Then:
2a^2 + b^2 = 1
4a^2 = (a*sqrt(2)-a)^2 + a^2 + (2b)^2
Substituing the first into the second and simplifying yields:
4a^2 = (sqrt(2)-1)^2*a^2 + a^2 + 4*(1-2a^2)^2
0 = (-8 - 2*sqrt(2))a^2 + 4
a = sqrt( 2 / (4+sqrt(2)) ) = sqrt( (4-sqrt(2)) / 7 )
The edge length equals 2*a, which is 2*sqrt( (4-sqrt(2)) / 7 ) = 1.21556 |
