We watched a pizza-cooking contest on the Food Network on TV. The master chefs who presented their creations cut their pizzas in an unusual way (see picture), creating 4 squares in the center, 8 slightly rounded pieces, and four smaller "triangle-like" pieces.

We started wondering where you need to cut the pizza in order to have the 12 larger pieces (numbered 1-12 above) be of the same size (here size means area).

Area shaded in yellow is^{ }_{ }

because^{ }_{ }

Notice that the area shaded in yellow contains two pieces. ^{ }_{ }One is a square with area c^{2}, and the other is a "rounded square" sitting on top of the square.^{ }_{ } We want the two pieces to have the same area, so the area outlined by dotted lines should have area 2c^{2}.^{ }_{ }
So we want to solve the equation _{0}∫^{c}√(1-x^{2})dx - 2c^{2} = 0 for c.^{ }_{ }

Solving the equation: _{0}∫^{c}√(1-x^{2})dx - 2c^{2} = 0 for the variable c.^{ }_{ }
First method.^{ }_{ }
Define,^{ }_{ }
\Y1=fnInt(√(1-X^{2}),X,0,C)-2C^{2}^{ }_{ }
Use SOLVER to solve the equation Y1 = 0 for variable C (not for X!).^{ }_{ }
The solution is almost exactly .48.^{ }_{ } It means that the radius is cut in the ratio of 48 to 52, or 12 to 13, almost in half.^{ }_{ }

Second method.^{ }_{ }
Use the transformation x = sin(t), then dx = cos(t) dt and c = sin(a).^{ }_{ } The equation is changed into: _{0}∫^{a}cos(t^{2})dt - 2sin(a)^{2} = 0.^{ }_{ } But _{0}∫^{a}cos(t^{2})dt = sin(2a)/4 + a/2^{ }_{ } (you may find it in tables of integrals, or on the internet at http://en.wikibooks.org/wiki/Calculus/Tables_of_Integrals, or derive it yourself).^{ }_{ } This reduces the problem to solving the equation sin(2a)/4 + a/2 - 2sin(a)^{2} = 0, for a.^{ }_{ } You may use SOLVER to do it, and then to compute c = sin(a).^{ }_{ } Also you may solve this equation using Newton's method on the TI-34 II calculator.^{ }_{ }

Remark.^{ }_{ }
The first method is much simpler, but you can use the second using a simpler calculator.^{ }_{ }

Task.^{ }_{ }
Can you divide this pizza into 16 pieces so that 12 of them have equal area, as outlined above?^{ }_{ }