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It's nearly pancake day here in the UK and I'll be making my kids pancakes. The last one in the batch is always a little oddly shaped, and I'd like to divide it into two pieces of equal area so they can share it. However, the children are fussy eaters and insist on their pieces being path-connected. Is this possible? Maybe there's an old theorem somewhere that would help me have a stress-free day?

(n.b. I'm capable of making very precise and wiggly cuts with the tip of the knife, so model the cut as a continuous map $C:[0,1] \to \mathbb{R}^2$)

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  • $\begingroup$ Is one of them hairier than the other? $\endgroup$
    – Lubin
    Feb 17 '17 at 16:28
  • $\begingroup$ I think for cutting in two, you can always find a straight cut, but I can't find the source at the moment. I'll post it as an answer when I find it. $\endgroup$
    – Simon Marynissen
    Feb 17 '17 at 16:36
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    $\begingroup$ @SimonMarynissen. For many non-convex pancake shapes, you can’t find a straight cut that yields two path-connected pieces. worldartsme.com/images/spiral-background-clipart-1.jpg $\endgroup$
    – Steve Kass
    Feb 17 '17 at 16:56
  • $\begingroup$ Yeah, I forgot the path-connected condition. $\endgroup$
    – Simon Marynissen
    Feb 17 '17 at 16:57
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If you have $2^n$ kids, you can use a generalization of the Ham sandwhich theorem to cut it using only $n$ cuts. In $2$ dimensions this is: if you've $m$ pieces of pancakes, there exists a curve that splits all the $m$ pieces in two pieces of the same size (area). So if you got $2^n$ kids, you start by dividing that piece in $2$ and then those two pieces each in $2$ with one cut. You can repeat this to get $2^n$ pieces.

Edit: I missed your stuff about path-connectedness. The theorems I used don't assure that.

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  • $\begingroup$ Yup, the path-connected condition is the tricky bit $\endgroup$
    – user33750
    Feb 18 '17 at 12:18
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If you are capable of making very precise and wiggly cuts then you should be able to divide the last batch precisely into two equal amounts before frying it.

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  • $\begingroup$ For someone so good at cutting, I am surprisingly bad at pouring ;) $\endgroup$
    – user33750
    Feb 18 '17 at 14:46

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