Cutting $n$ circular cakes of different radii but equal heights into $p>n$ equal shares

Question

You probably know the following problem:

We have two circular cakes of the same height but unknown and potentially different radii, and we want to cut them into two equal shares. Each cut can only cut one piece of one cake. What is the minimal number of cuts required?

The solution is easily found:

It's $1$ cut. Stack the two cakes and center them, then cut the big one around the small one : Here you go.

While I was thinking about this easy problem, the following question logically arose:

We have $n\in\mathbb{N}^*$ circular cakes of unknown and potentially different radii (hmm yummy) but the same height, and we want to cut them into $p\gt n$ equal shares. Each cut can only go through one piece of cake. What is the minimal number $c$ of cuts required? How do you cut the cakes?

The cuts can be done however you want (i.e. lines, curves, etc) and stacking cakes etc is allowed. You are of course not allowed to measure the exact radius. To make it simple, let's say you have a straightedge (unlabelled ruler), an arbitary-precision protractor and a compass.

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Any help/thoughts are appreciated.

Answer 1

Proof that you need at most $p - 1$ cut:

It is known that using compass and straightedge, with a segment of length $a$ and $b$, $a \geq b$, one can construct $c = \sqrt{a^2+b^2}$ and $d = \sqrt{a^2-b^2}$. It is also possible to divide a segment into an arbitrary number of equal parts. Addiontially, the center of a circle can be determined. This is sufficient for us to construct the fair cuts.

Let $r_1,r_2,...,rn$ be the radii of the circles. Let $r^{*}$ be the radius of the circle whose area is equal to the area of a share, then $$r_* = \frac{1}{p}\sqrt{r_1^2+r_2^2+...+r_n^2}$$ which is clearly constructible. Now let's put the $n$ circles into the first $n$ shares, as the same time we also draw $p$ segments to keep the areas of the shares of length $x_1,x_2,...,x_p$ in check (for convinience we call them the shares' radii). Initially, $x_i = r_i$ for $i \leq n$ and $x_i = 0$ for $i > n$. We proceed to repeat the following steps

Find a largest share with radius $x_\max$ and a smallest share with radius $x_\min$. If $x_\max > r'$, proceed to cut and remove an annulus piece of area $S_{cut} = \pi x_\max^2 - \pi r_*^2 = \pi r_{cut}^2$ then move it to the smallest share and update their radii accordingly. Otherwise, halt. For an annulus with radii $a,b$, $a < b$, to cut a piece of area $S_{cut}$, we have

$$r'_{cut} = \sqrt{r_{cut}^2 + a^2}$$

After each operation, the number of share that has the radius equal to $r_*$ increases by at least $1$. Since there're only a finite number of shares, eventally the process halts. The possible scenarios are

• One cannot find a piece large enough. That's impossible because the largest piece of a share is either the initial cake, or a piece that it receive from a previous step. However, in that case, the surplus area received is $S_{sur} = S_{cut} + S_\min - S_* < S_{cut}$ therefore the surplus that's needed to be cut off cannot be larger than the largest part (in terms of area).

• $x_{\max} = r_*$. This only happens when all the shares are equal

Thus our process of cutting the cakes give the equal shares as desired. Since at each step at least $1$ equal share is produced, and only $p-1$ equal shares is needed, since the leftover is obviously also equal, only at most $p-1$ cuts are needed.

P/s: I do not know how to prove if you need at least $p-1$ cut for almost all sets of cakes. Someone who is well versed in number theory may be able to prove that.