A, B, and C are points on a sphere; if the spherical coordinates of A are (1, pi/2, 0), and ...?

Question

the spherical coordinates of B are (1, pi/3, pi/3), then what
are the possible spherical coordinates of point C, such that
the points A, B and C form an equilateral triangle.

Note that A, B, and C are on the same sphere. so the first
coordinate of C is equal to 1 (the radius of the sphere).

Also note that the second coordinate above is (theta), the angle
between the position vector of a point and the z-axis.  And the third coordinate (phi) is the angle between the projection of the position vector onto the xy-plane and the x-axis.

Indica · Accepted Answer

A = (1,π/2,0), OA = (1,0,0)  :  B = (1,π/3,π/3), OB = ( (√3)/4, 3/4, 1/2 )  
For C = (1,θ,ϕ), OC = ( sinθcosϕ, sinθsinϕ, cosθ ). OA, OB, OC are unit vectors.

For ABC equilateral angle subtended by AB, BC, CA at centre are the same

∴ OA•OC = OA•OB and OB•OC = OA•OB

These lead to
 
sinθcosϕ = (√3)/4    …   (i)

sinθcosϕ + (√3)sinθsinϕ + (2/√3)cosθ = 1      …   (ii)

Using (i) in (ii) gives sinθsinϕ = 1/√3 − 1/4 − (2/3)cosθ   …   (iii)

(i)²+(iii)²   →   sin²θ = 3/16 + ( 1/√3 − 1/4 − (2/3)cosθ )²

As a quadratic in cosθ this is 52cos²θ + (12−16√3)cosθ − (15+6√3) = 0

Wolfram gives cosθ = (√3)/2, −(6+5√3)/26

Since 0≤θ≤π this gives θ=π/6, θ ≈ 2.1698436322

ϕ ∈ [0,2π] comes from (i) resp. as ϕ=π/6, 11π/6 : ϕ ≈ 1.0188929489, 5.26429235828

Squaring process introduces the possibility of extraneous solutions so check.

The ones that work are θ=π/6, ϕ=11π/6  :  θ=2.1698436322, ϕ=1.0188929489

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You could also do it by solving for the intersection of 3 spheres.

The side of the planar triangle ABC is from s²  = (1−√3/4)²+(3/4)²+(1/2)²

(x−√3/4)²+(y−3/4)²+(z−1/2)² = s²

(x−1)²+y²+z² = s²

x²+y²+z² = 1

Having solved these (and x is easy) you then solve the trig to get sphericals

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A, B, and C are points on a sphere; if the spherical coordinates of A are (1, pi/2, 0), and ...?

1 Answer