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?

  • $\begingroup$ Lazy, but fair. $\endgroup$ – 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.

Similarly, 4 cuts won't work: 2 of the lines cut off segments with area of 5/11 of the circle, and 2 cut off segments with area 4/11. Again, you can compute the distance each line must be from the center of the circle, and guarantee that there will be pieces near the center that will be too small.

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.


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