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Solution of the Week #499 - Octagon to Square

December 30, 2024 Elliott Line

The above shows how the dissection would be achieved. If we call the side length of the octagon s and the radius of the small circles r, then we can say that an octagon with side s, subtract an octagon with side 2r must have the same area as a square with side 2s.

The formula for the area of a regular octagon is (2+2sqrt2)s^2. The area of the inner octagon is 4(2+2sqrt2)r^2, and the area of the full square is 4s^2.

 (2+2sqrt2)s^2 - 4(2+2sqrt2)r^2 = 4s^2

 Let’s use this to find a relationship between s and r.

 r = s/2(sqrt(3-2sqrt2))

 If we let the square be a unit square, s is 1/2 and therefore r is (sqrt(3-2sqrt2))/4. The area of each of the circular holes is pi*r^2 = pi*(3-2sqrt2)/16.

 The shaded part of the square is 1 less the area of five holes,

1-5*pi*(3-2sqrt2)/16

which is approximately equal to 0.83156…

So the square is about 83.156% shaded.

← Solution of the Week #500 - RectangleSolution of the Week #498 - Three Week Odyssey →

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