Here's my approach.....for simplicity.....I set the radius = 1.......we can adjust for this in the final answer....
Lett DE and BC be perpendicular diameters......now imagine that one of the cuts will be made along chord FH parallel to BC......and connect the center of the circle, at A with F and let point I lie on BC such that IF ll AG
The area bounded by the irregular shape BFGA [ rectangle IFGA and the area between the semi-chord IF and the minor arc BF ] will be (1/4) of (1/3) of the area of the pizza = (1/3)* (1/4) pi * r^2 = pi/ 12
And let angle BAF = θ......and because IF and AG are parallels cut by a transversal, angle FAB = angle GFA
And the area of the sector ABF = (1/2)r^2* θ = (1/2) (1)^2 * θ = θ/2
And the area of right triangle AGF = (1/2) AG * GF = (1/2) * (r)sinθ * (r)cosθ = (1/2)* (1)* sinθ* (1)*cosθ = (1/2)sinθcosθ
And these two area comprise the total area BFGA
So we have that
θ/2 + (1/2)sinθcosθ = pi/12 multiply through by 2
θ + sin θ cos θ = pi / 6
Using Wolframalpha to solve → θ ≈ 0.268133
And the distance from the center to the cut is given by r* sin (0.268133) = r * 0.264932 = 0.264932* r
Just as Melody found !!!!