Since the octagon is cyclic, the sides can be arranged in any order without affecting the values of a, b, or r. Rearranging the sides in alternating order b, *a*√2, b, *a*√2, b, *a*√2, b, *a*√2 creates a symmetric equiangular octagon, equivalent to a square with sides 2a+b with the corners truncated by length a.

Draw two lines from the center, one to a vertex and the other as a perpendictular bisector of a b length side. The triangle created has legs of b/2 and (2a+b)/2 and a hypotenuse of r. Integer solutions can be created very easily by taking any integer Pythagorean triple and mapping the smallest member to b/2, the second to (2a+b)/2 and the largest to r.

PT(3,4,5) -> b=6, a=1, r=5 (smallest b>a)

PT(7,24,25) -> b=14, a=17, r=25 (smallest a>b)