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What percentage of the height is the distance from the vertex of a regular tetrahedron to the center of gravity? Tell me the general meaning.

The volume of a regular tetrahedron (triangular pyramid) is one third of the bottom area times the height.

v0= 1/3*s0*h

The height above the center of gravity is x.

v 1 = 1/3s 1 * x s 1/s0=(x/h)^2

v 0 = 2v 1v 1/v 0 =( 1/3 * s 1 * x)/( 1/3 * s0 * h)

=s 1/s0*(x/h)=(x/h)^3

(x/h)^3= 1/2

x=( 1/2)^ 1/3≈0.7937