## Friday, January 7, 2011

### Circular cutouts

Find the ratio of the shaded area to the area of the square.  No calculus is required.

This is not an original puzzle.  Rather, I consider it a classic.

see the solution

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

Those are perfect quarter-circles, correct?

miller said...

Correct.

Eduard said...

pi/3 + 1 - sqrt(3)

Eduard said...

(1) : b + 2c = 1 - pi/4
(2) : a + 2b = pi/2 - 1
(3) : a + b2 + c = pi/3 - sqrt(3)/4

From (2) and (3) we get c = 1 - sqrt(3)/4 - pi/6

From (1) we then get b = 1 - pi/4 - 2c = pi/12 + sqrt(3)/2 - 1

From (2) we then get a = pi/2 - 1 - 2b = pi/3 + 1 - sqrt(3)

Proof of Eqn (1): The region consisting of one b and its two neighboring c's is shaped like a "concave triangle" wedged into one corner of the square. It is obtained by cutting away a quarter circle from the square.

Proof of Eqn (2): The region consisting of the central a and two opposite b's is shaped like a "lens". It is obtained by "painting" the areas of two opposite quarter circles (thereby painting the lens *twice*, then "unpainting" the area of the square itself. (This procedure would leave only the lens painted, with a single coat of paint.)

Proof of Eqn (3): The region consisting of the central a, one c, and the two b's that are adjacent to that c, is shaped like an equilateral triangle two of whose sides bulge out (the third side being flat, being a side of the square).
To find the area of this, we slice off one of the bulging edges. This breaks the area up into two pieces, called X and Y, say. X is an equilateral triangle with one bulging arc-shaped side. Note that this is exactly the shape one gets by cuttign a circle into 6 congruent pieces by radial cuts. So X = pi/6.
At the same time, Y is a "half-lens", with one flat side and one bulging side. We note that Y can be obtained by taking an X-shaped area, and removing the large equilateral triangle from it. So Y = pi/6 - sqrt(3)/4.

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