# Cutting the square

I don't know how simple/diffcult this is, but it has been on my mind for some time, and I haven't managed to do anything about it, so now I'll try here to see if anyone has any input.

Let a natural number $n\in\mathbb N$ be given.

Start with a square $S_0$ with sidelength $1$.

For $k$ in $\{1,2,\ldots, n-1\}$, define $B_k$ as a section of $S_{k-1}$ of area $\frac{1}{n}$ defined by a straight line, and $S_k$ as $S_{k-1}\setminus B_k$. (I believe a simple continuity argument proves the existence of $B_k$.)

And set $B_n = S_{n-1}$.

(If anything in the process is unclear, it might help to know it was inspired by cutting a slice of bread, $S_0$ is the initial slice, and the $B_k$'s are "bites" of equal size $1/n$)

What is $\inf\left\{\sup\left\{\operatorname{diam}(B_k)\mid k\in \left\{1,2,\ldots,n\right\}\right\}\mid\text{every set of }B_k\text{'s}\right\}$?

(I don't find it easy to parametrise the set of $B_k$'s, but I hope it's understandable. - To continue the explanation of how I came up with the problem, this question is (an attempt of) a formalisation of "What is the smallest the biggest "bite" can be?")

(Where $\operatorname{diam}$ is the diameter of a set, defined as $\operatorname{diam}(A) = \sup\{d(x,y)\mid x,y\in A\}$ - I think that's standard in metric spaces)

In the faint hope it helps someone, here are my thoughts for the simple cases:

For $n=1$: There will only be one "bite", being the whole square, so the answer is the diameter of that, or $\sqrt{2}$.

For $n=2$: There will be two "bites", the cut between them will go through the centre of the square and the "bites" will be similar. The diameter will be between $\sqrt{2}$ (for the "bites" being isoceles triangles, the cut going from one corner of the unit square to the opposite) and $\sqrt{5}/2$ (for the "bites" being rectangles, the cut going from the midpoint of one side to the midpoint of the opposite side). The answer to my question being $\sqrt{5}/2$.

For $n\geq 3$: I think the first "bite" (this is actually similar to the case of $n=2$, but in that case we were done afterwards) will vary between the isoceles right triangle with side $\sqrt{2/n}$ and a rectangle with one side having length $1$ (being the complete side of the square) and the other side having length $1/n$. The diameter of those figures are $\sqrt{4/n}$ and $\frac{\sqrt{n^2+1}}{n}$ (for $n=2$ this agrees with the figures we got above). If we rewrite those expressions a little they become $$ \sqrt{\frac{4n}{n^2}} \text{ and } \sqrt{\frac{n^2+1}{n^2}} $$ for $n\geq 4$ the second of those are the biggest)

In between those two is the case where the cut goes through one of the corners of the square and forms a right triangle with one side being the side of the square. In that case the short side of the triangle has length $\frac{2}{n}$, giving the "bite" a diameter of $\sqrt{\frac{n^2+4}{n^2}}$ - for $n\geq 3$, this is bigger than both of the numbers above, so the supremum is assumed somewhere between the end points, but my guess is that the diameter of the isoceles triangle can be shown to be the infimum, making it the "smallest first bite".

But in any case we're left with several possibilities for the second cut and a figure that is not as easy to handle as the unit square (there's a possibility that a similar "bite" starting from another corner, might be minial. And as far as I can see it only gets worse - and there's no guarantee that the smallest first "bite" will allow the construction that attains the solution?

## 3 comments

OK, so $S_0$ is a unit square. $B_1$ is a section of $S_0$ of area $1/1$, that is, all of $S_0$ except for a null set. Since it is a section by a straight line, that null set can be either empty, a corner point or a side. $S_1 = S_0\setminus B_1$ (your $S$ at that point misses an index, but up to now, $S_0$ is the only $S$ already defined, so it must be that), so $S_1$ is that null set. So now you're down to a line segment at best. I strongly suspect that is not what you had in mind. — celtschk 5 months ago

No. $B_1$ does not have area $1/1$, $n$ is fixed throughout the procedure so all the $B_k$'s have area $1/n$., and I thought that was conveyed by using $k$ as index on the $B$'s when defining those. (And of course it's a mistake that I forgot an index on that $S$, I will add it right after posting this comment.) — Grove 4 months ago

For $n=3$ you want three rectangles of 1/3 by 1. Beyond there it gets more complicated; I suspect that the initial cuts will tend to leave a rough circle, but if so then IIRC some calculations I made a few months ago showed that the diameter of a sector sliced from the circle would eventually be greater than the diameter of the initial corner, which partially undermines hope for a greedy approach. You're correct that the isosceles triangle is the infimum for triangles cut from a corner. — Peter Taylor 3 months ago