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Calculating the exact coordinates of a regular tetrahedron circumscribed by the unit sphere at the origin

First, draw the first point O at the origin (0, 0):

Then, draw the second point A at (1, 0):

Then, draw the third point B so that △BOA is a equilateral triangle.

Since ∠BOA = 60°, and OA = OB = 1:

B = (cos 60°, sin 60°)
  = (1/2, √3/2)

Let point C be the centroid of △BOA. The coordinates of point C could be found be "averaging" the coordinates of O, A, and B.

C = ((0 + 1 + 1/2) / 3, (0 + 0 + √3/2) / 3)
  = ((3/2) / 3, (√3/2) / 3)
  = (1/2, 1/(2√3))

By extending the 2D space to 3D, here are what we got so far:

O = (0,   0,       0)
A = (1,   0,       0)
B = (1/2, √3/2,    0)
C = (1/2, 1/(2√3), 0)

Since the forth point D is "above" point C, let D = (1/2, 1/(2√3), d). Since OA = OB = AB = OD = AD = BD = 1, we take OD = 1:

OD² = (1/2)² + (1/(2√3))² + d²
  1 = 1/4 + 1/12 + d²
    = 3/12 + 1/12 + d²
    = 4/12 + d²
    = 1/3 + d²
2/3 = d²
  d = ±√(2/3)

By picking the positive root for easier calculation, we have D = (1/2, 1/(2√3), √(2/3)).

Let point E be the center of the tetrahedron. Since point E is on the line CD, let E = (1/2, 1/(2√3), e). Since OE = AE = BE = DE, we take OE = DE:

                      OE = DE
                     OE² = DE²
(1/2)² + (1/(2√3))² + e² = 0² + 0² + (e - √(2/3))²
                1/3 + e² = e² - 2√(2/3) e + (2/3)
                     1/3 = -2√(2/3) e + (2/3)
                    -1/3 = -2√(2/3) e
                     1/3 = 2√(2/3) e
                 1/(2*3) = √(2/3) e
                       e = √3/(2*3*√2)
                         = 1/(2√6)

So, here are what we got so far:

O = (0,   0,       0)
A = (1,   0,       0)
B = (1/2, √3/2,    0)
D = (1/2, 1/(2√3), √(2/3))
E = (1/2, 1/(2√3), 1/(2√6))

Since we need the vertices to be on a unit circle, the distance from the center to a vertex must be 1. That means, we must scale the points so that OE = AE = BE = DE = 1. Let k be the scaling factor:

                                 k * OE = 1
                               k² * OE² = 1
k² * [(1/2)² + (1/(2√3))² + (1/(2√6))²] = 1
               k² * (1/4 + 1/12 + 1/24) = 1
              k² * (6/24 + 2/24 + 1/24) = 1
                            k² * (9/24) = 1
                             k² * (3/8) = 1
                                     k² = (8/3)
                                      k = (2√2)/(√3), since k is positive

So:

O' = (0, 0, 0)
A' = (2√(2/3), 0, 0)
B' = (√(2/3), 2, 0)
D' = (√(2/3), √2/3, 4/3)
E' = (√(2/3), √2/3, 1/3)