A square both is inscribed by a circle and inscribes a circle. Exactly halfway between these two circles lies a third circle. This third circle is inscribed by a regular octagon.
Which polygon has a larger perimeter–the square, or the octagon?
The answer is…(Highlight for the solution)
Let the inner circle’s radius equal r1. Given the geometry of the square, the radius of the outer circle must equal r1*√2 (since that is the distance from the center point of the square to its vertices). Thus, the middle circle must have a radius of the average of the two other circles, r1*(1+√2)/2. The height of the octagon is twice this value, r(1+2^.5).
Let the value of an edge of the octagon be some value s. In terms of s, the height of the octagon is s + s/√2 + s/√2, or s(1+√2). By setting this value for the octagons height equal to the original expression of the octagon’s height, we can relate r to s: r(1+2^.5) = s(1+2^.5). Thus, r = s.
Since the edge length of the square is 2*r, its perimeter is 4 times this value, or 8*r. The perimeter of the octagon is simply 8*s. Since r = s, these are the same quantities. Neither perimeter is larger.