(1) Cuts parallel to each other.

Consider the above circle of radius r centered at the origin O = (0, 0) where the two parallel cuts are AB and CD. We note the center cut ABCD is exaggerated in size, but we do so for effect only.

If q (theta) is the angle BOP then we shall determine the area of the center region by doubling the top half above the x-axis. Further, we consider symmetry and the upper region of the circle above line AB is presumed to be the same size as the lower region of the circle below CD.

Thinking in a rectangular coordinate system we note that the line y = tan(q) x from O to B is significant as are the coordinates of the P = (r cos(q), r sin(q)) - here q is the angle theta in the diagram.

Thus the area of middle section ABCD is two times the sum of the areas of the four congruent triangles BOP, EOB, AOE, AOQ and the two pieces BPM and AQN.


In our analysis we use t for the angle q (theta)

We simplify a bit and determine where this MidPiece area is 1/3 Pi r^2.

Thus the angle which we must use for q (theta) is 0.268 radians.

We confirm the area of the MidPiece IS 1/3 Pi r^2 = 1.0472 r^2.

And we check the top piece to see if it is one third the area of the circle, or 1/3 Pi r^2 = 1.0472 r^2. It is!

Hence we should make two horizontal cuts at heights TopCut and BottomCut respectively.

These two cuts will give us a top, a middle, and a bottom piece each of 1/3 the total area of the cake.

Of course we could have used polar integration to obtain the area of the sector of the circle BOM (or even the formula for the area of a sector of a circle) and then proceeded to add in the triangular areas. This leads to the same result.