You have a special eight-bladed pizza cutter. All you do is pick a point on the pizza, and the device cuts out eight straight lines from that point to the circumference of the pizza, separated by equal 45 degree angles.

You and your friend just bought a pizza and would like to have four slices of pizza each. Your friend tells you that you can make the cut using your device, using any center point you would like. After the cuts have been made, the two of you will eat alternate slices (so that nobody eats two adjacent slices).

How much of the pizza can you end up with?

(In reply to re: thoughts by Richard)

The somewhat messy approach works!

Taking Charlie's formula

D = D(theta) = cos(theta) + sqrt((cos(theta))^2 + a^2 -1)

as the distance from the center of the pizza cutter to the edge of the pizza, we can calculate the area A=A(T) of the piece of the pizza that that has its cut edges at theta = T + k*pi/4 and theta = T + (k+1)*pi/4  by integrating the square of D over theta from T + k*pi/4 to T + (k+1)*pi/4, where k is an integer.  The rate of change of the area of such a piece of the pizza with respect to T is thus

[D(T + (k+1)*pi/4)]^2 - [D(T + k*pi/4)]^2.

The rate of change with respect to T of the sum S of every other piece can then be written as

S = [D(T + pi/4)]^2 - [D(T)]^2 + [D(T + 3*pi/4)]^2 - [D(T + 2*pi/4)]^2 + [D(T + 5*pi/4)]^2 - [D(T + 4*pi/4)]^2 + [D(T + 7*pi/4)]^2 - [D(T + 6*pi/4)]^2.

We have

[D(X)]^2 = [cos(X)]^2 + 2*[cos(X)]*sqrt((cos(X))^2 + a^2 -1) + [cos(X)]^2 + a^2 - 1

            = 2*[cos(X)]^2 + 2*[cos(X)]*sqrt((cos(X))^2 + a^2 -1) + a^2 - 1

            = cos(2*X) + 2*[cos(X)]*sqrt((cos(2*X))^2 + a^2) + a^2

and we may ignore the final a^2 as it washes out in S.

Employing the formula for the cosine of a sum, we evaluate each of the terms of S and find that the terms with radicals are identical but with opposite signs for offsets of T that are pi apart, and the other terms are identical but with opposite signs for offsets of T that are pi/2 apart. Hence S vanishes for all T and all a, and therefore the sum of the areas of every other piece of pizza made by the cutter, no matter where its center is placed on the pizza, is a constant.

Posted by Richard on 2005-02-09 05:51:47