The smallest size of the squares is 1 + (sqrt(6) - sqrt(2))/4 = 1.25882.
Each square covers one third of the circumfrence of the circle and two sides of each square are tangent to the circle.
Let ABCD be one of the squares and let the circle have center O. E (on AB) and F (on BC) are where the square is tangent to the circle. G (on CD) and H (on DA) are where the square intersects the circle. I is where an extension of EO intersects CD.
Since arc HG equals 120 degrees (one third of the circimfrence) the angle HOG is 120 degrees.
Using the law of cosines GH = sqrt(3).
Triangle HDC is a 45 degree right triangle, so then HD = sqrt(3/2).
Angle HOB is 60 degrees (half angle HOG), then angle HOD is 120 degrees.
Using the law of cosines OD = (sqrt(3) - 1)/2.
Then OI = (sqrt(3) - 1)/sqrt(2) and AD = EI = EO+OI = 1+(sqrt(3) - 1)/sqrt(2) = 1 + (sqrt(6) - sqrt(2))/4. |