Harmonic Ratio of the Octahedron, ~1.7321

From Euclid XIII.14 (Heath, 1960; v.3, p. 474) we have,
(2*R)2 = 2*S2, or
[1] S = sqrt(2)*R
Again, to relate S and r (and thus R and r) we will resort to the central triangle.

The central triangle argument for the octahedron

Again, this triangle connects the center of the octahedron to a vertex, the side R, to the center of a equilateral triangular face, the side r, and across the triangular face, side c. Applying the Pythagorean theorem to this triangle, we have as in each case,
[2] R2 = r2 + c2,
In this case, again, c is the distance from the center of a equilateral triangle to one of its vertices. This distance is related to S, the edge of the triangle, by
[3] S = sqrt(3)*c
as we have seen several times before. So now, substituting [3] in [1] and squaring, we find,
[4] c2 = S2/3 = (2/3)*R2
Finally, putting this expression in place of c2 in equation [2] and collecting terms in R2 on the left, we have,
[5] (1/3)*R2 = r2
so the harmonic ratio we seek is,
[6] R/r = sqrt(3)
which evaluates to 1.732051, rounded to the sixth decimal place as usual.