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The ham sandwich theorem states that given $n$ measurable "objects" in $n$-dimensional space, it is possible to divide all of them in half (with respect to their measure) with a single $(n−1)$-dimensional hyperplane.

In $n$-dimensional Euclidean space, can we divide $n$ objects into thirds using two hyperplanes of codimension $1$? More generally: can we divide each object into $k$ parts of equal volume using $k-1$ hyperplanes?

I know these $k-1$ hyperplanes would not necessarily be disjoint (visualize a huge object and a tiny object a small distance apart in $\Bbb{R}^2$ and try slicing them $k$ times--the lines clearly intersect), in which case I ask: What properties must the objects have in order for the hyperplanes of codimension $1$ to be disjoint?

It seems like you can divide into $k$ parts by the hyperplane separation theorem. The method of division seems easier when $k$ is even, and not so obvious for odd $k$.

Along with the method for dividing into $k$ (odd) equal parts, the question I would really like an answer to is: when are the hyperplanes of codimension $1$ disjoint?

Edit: since this question isn't getting as much attention as I hoped, I'd like to revise the bounty request to simply any insight on the questions I have--a full proof is not necessary if you cannot offer one. Feel free to just share your ideas about this or start a discussion! Thanks!

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  • $\begingroup$ Crossposted; mathoverflow.net/questions/201854/… $\endgroup$ – Qiaochu Yuan Apr 3 '15 at 3:26
  • $\begingroup$ To help your exploration, you can't do it in $R^2$ and divide into fifths with four lines. Suggested by san, let $X$ be two small discs and $Y$ be a large annulus around them. One of the lines can go between the discs of $X$, but the others have to go through them. That will give you four big pieces of $Y$. $\endgroup$ – Ross Millikan Apr 8 '15 at 3:43
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You can have certainly a line cutting two objects in $R^2$ at a third and two-thirds each. Then you can construct another line cutting the two two-third parts in half. The problem is that the two lines may intersect, so they can cut one of the objects in more parts (four).

Take for example $X=D_1(0,5)\cup D_1(0,-5)$ and $$ Y=D_1(5,0)\cup D_1(10,0)\cup D_1(15,0)\cup (\bar D_{10}(10,0)\setminus D_{10}(10,0)) $$ Here $D_r(x,y)$ is a disk with center $(x,y)$ and radius $r$. This means, $X$ is formed by two small disks and $Y$ by three disks and a (big) circle.

Any two lines such that the first cuts simultaneously $X$ and $Y$ in a big part that is the double of the other part, and such that the second one cuts $X$ the big parts in halfes, necessarily cuts $Y$ in four parts.

In $R^3$ it is even worse: Take $X=B_1(0,0,10)$, $Y=B_1(0,0,-10)$ and $Z$ a torus around the circle $x^2+y^2=10^2$, $z=0$; say with smaller radius 1. Any plane that passes through $X$ and $Y$ cannot cut the torus $Z$ in a part that is the double of the other.

I'm sorry, my first assertion is false, if you take a small disk and a big anulus with the same center, then you cannot have a line cutting the two objects in $R^2$ at a third and two-thirds each simultaneously.

If you are allowed to cut each object in four pieces, then the result (in $R^2$) seems to be true.

Here an (incomplete) approach: Take all the lines that cut an object in one-third, they are a continous family of lines going around. If none of the lines chops off the other object a third, then either it chops off something that is always smaller than one third, or always greater than one third. Changing eventually the objects one can assume it is always greater. Fix one such line, which cuts the second object in parts A,B; B being on the same side as the one third part of the first object. Then the lines that cut the two-third part in two halfes are a continuous family of lines that cut $A$ in $A_1, A_2$ and $B$ in $B_1, B_2$. One of them satisfies that $B_1+A_1$ (opposite pieces) is one third.

Now it suffices (and this is the incompleteness) to prove that moving around the fixed line (and the whole construction) you get at some point the equality $A_2=B_2$ (equal to one third).

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You can't do this. In $\Bbb R^2$ consider three small circles at the corners of a triangle. Each line only intersects two of the circles, so of two lines there will either be two circles that only meet two lines or one circle that doesn't meet any lines. Clearly not all circles will be divided into thirds.

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    $\begingroup$ The original theorem seems to require the number of objects to be the dimension of the space. $\endgroup$ – Ben Blum-Smith Apr 3 '15 at 4:33
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    $\begingroup$ Isn't this one object too many? Otherwise this would be a counterexample to the original theorem too... $\endgroup$ – Steven Stadnicki Apr 3 '15 at 4:33
  • $\begingroup$ This shows that we must have $k \le n$ in the more general question. Working on the rest. $\endgroup$ – Ross Millikan Apr 3 '15 at 4:36
  • $\begingroup$ But must we have $k\leq n$? I think any $k$ works. My idea was to consider the two disjoint convex sets of Euclidean space created by the first "ham slice" hyperplane. Then we have $n$ half-objects which are in a convex $n$-dimensional space, so we can just take another hyperplane "ham slice" in each set (and repeat). I cannot think of such a method when $k$ is odd, though, which is partially why I asked about dividing into thirds in the title. $\endgroup$ – Patrick Shambayati Apr 3 '15 at 4:43
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    $\begingroup$ I revised the question to "In $n$-dimensional space ... $n$ objects". I am aware that $n+1$ objects in $\Bbb{R}^n$ fails $\endgroup$ – Patrick Shambayati Apr 3 '15 at 4:55

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