A unit equilateral triangle will have an inradius of 1/(2*sqrt[3]).  Place the triangle on a coordinate grid with the center at the origin and a vertex on the positive x-axis; let the vertex on the x-axis be the vertex corresponding to length b.  The incircle will have a formula of x^2 + y^2 = 1/12 and the vertex on the x-axis is at (1/sqrt[3], 0).

A general point on the incircle will have coordinates ((cos t)/(2*sqrt[3]), (sin t)/(2*sqrt[3])).  The distance from this point to the vertex on the axis is b = sqrt[(cos t - 2)^2/12 + (sin t)^2/12] = sqrt[(5 - 4*(cos t))/12].

When the triangle is rotated 120 degrees clockwise from its initial configuration then the second vertex is on the x-axis.  The distance of this vertex to the point on the incircle is a=sqrt[(5 - 4*(cos (t-120)))/12].

Similarly, when the triangle is rotated 120 degrees counterclockwise from its initial configuration then the third vertex is on the x-axis.  The distance of this vertex to the point on the incircle is c=sqrt[(5 - 4*(cos (t+120)))/12].

For a, b, and c to be in geometric progression then b^2 = a*c, which means (sqrt[(5 - 4*(cos t))/12])^2 = sqrt[(5 - 4*(cos (t-120)))/12] * sqrt[(5 - 4*(cos (t+120)))/12].

Eliminating radicals and dividing out common factors yields [5 - 4*(cos t)]^2 = [5 - 4*(cos (t-120))] * [5 - 4*(cos (t+120))].

Expanding and simplifying each side reduces the equation to 25 - 40*(cos t) + 16*(cos t)^2 = 25 + 20*(cos t) + 4*(cos t)^2 - 12*(sin t)^2.

Then a last round of simplification reduces the equation to cos(t) = 1/5.  This then makes b = sqrt[(5 - 4*(1/5))/12] = sqrt[7/20] = sqrt[35]/100.

Then a = sqrt[(5 - 4*((cos t)*(-1/2) - (sin t)*(-sqrt[3]/2)))/12] simplifies to a = sqrt[(5 - 4*((1/5)*(-1/2) - (sqrt[24]/5)*(-sqrt[3]/2)))/12] 

= sqrt[(5 + 2/5 - 12*sqrt[2]/5)/12] 

= (2*sqrt[2] - 1)/(2*sqrt[5]) = sqrt[10]/5 - sqrt[5]/10.

Similar calculation shows c = sqrt[10]/5 + sqrt[5]/10.