## Sunday, February 6, 2011

### Solutions to circular cutouts

See the original puzzle

Above, I've highlighted several regions which we will use as intermediate steps to find the area of the center.  The letters A through D represent the area of that region.  I'm taking the square to have unit length sides.

A is a sixth of a circle.
A = 1/6 pi

B is an equilateral triangle.
B = sqrt(3)/4

C can be found by combining A and B.
C = 2A - B = 1/3 pi - sqrt(3)/4

D is a quarter of a circle.
D = 1/4 pi

By tinkering around with C and D, we can find the area of the center region.
Area of center = 1 - 4D + 4C = 1/3 pi + 1 - sqrt(3)

Larry, a.k.a. The Barefoot Bum said...

C can be found by combining A and B.
C = 2A - B = 1/3 pi - sqrt(3)/4

???

By tinkering around with C and D, we can find the area of the center region.
Area of center = 1 - 4D + 4C = 1/3 pi + 1 - sqrt(3)

???

miller said...

Take area A and overlay it on its own mirror image. This creates region C, except that a triangular-shaped region has been counted twice. So if you subtract off the area of the triangle, you get the area of C.

The second formula can be understood in the same way, but it's more involved.

Larry, a.k.a. The Barefoot Bum said...

It would probably help if I had even a little capacity at visualizing shapes. I'm pretty strictly an algebra thinker, not a geometry thinker.