# Equal areas of segments in the lazy caterer problem?

In the book "Build Your Brain Power" by Wootton and Horne, they mention the lazy caterer's problem, asking for a way to cut a circular cake into 8 equally sized pieces with 3 cuts. Clearly since the maximum number of possible segments is 7 for the $n=3$ case, that is impossible.

But, I was nonetheless wondering about this: is it possible, for any $n\geq 3$ (3 cuts or more) case of the lazy caterer's problem, for each of the resulting areas to be equal? How would one prove or disprove this?

• Lazy, but fair. – Eric Tressler Dec 19 '15 at 4:11

Well, it's not possible for three cuts. Each cut has 3 pieces on one side of it, and 4 on the other, so it must cut off 3/7 of the circle on the side away from the center. That allows you to solve for the distance of the cut from the center of the circle; it turns out to be about 1/9 of the radius. Since all three cuts must be the same distance from the center, and since the three pieces between the acute angle formed by a pair of cuts and the circle must all have the same area, the acute angle between any pair of cuts must be the same, and so must be $$\tau/6$$ ($$\tau$$ being the angle measure of a full circle). So the central piece must be an equilateral triangle centered on the center of the circle, with height roughly 1/3 the radius (3 times the distance from the center to each side). But unfortunately the area of such an equilateral triangle is much less than 1/7 the area of the circle.
Presumably, the situation just worsens from there, and there will always be too-small slivers near the center, but this presumption falls short of a proof that there is no equal-area lazy caterer dissection for any $$n$$. But I'd be astonished if there were.