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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!

Patrick Shambayati
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  • Crossposted; http://mathoverflow.net/questions/201854/ham-sandwich-theorem-extended-to-divide-into-k-parts – Qiaochu Yuan Apr 03 '15 at 03:26
  • 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 $Y$. – Ross Millikan Apr 08 '15 at 03:43

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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.

Ross Millikan
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    The original theorem seems to require the number of objects to be the dimension of the space. – Ben Blum-Smith Apr 03 '15 at 04:33
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    Isn't this one object too many? Otherwise this would be a counterexample to the original theorem too... – Steven Stadnicki Apr 03 '15 at 04:33
  • This shows that we must have $k \le n$ in the more general question. Working on the rest. – Ross Millikan Apr 03 '15 at 04:36
  • 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. – Patrick Shambayati Apr 03 '15 at 04:43
  • I think my argument will extend to $n+1$ hyperballs at the corners of a simplex in $\Bbb R^n$ – Ross Millikan Apr 03 '15 at 04:47
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    I revised the question to "In $n$-dimensional space ... $n$ objects". I am aware that $n+1$ objects in $\Bbb{R}^n$ fails – Patrick Shambayati Apr 03 '15 at 04:55