Alice and Bob are sharing a triangular pie. Alice will cut the pie with one straight cut and pick the bigger piece, but Bob first specifies one point through which the cut must pass. What point should Bob specify to get as much pie as possible? And in that case how much pie can Alice get?

The approach I took was to put an arbitrary point in a triangle, and draw lines from it to each corner of the triangle. Now the cut will only go through two of the three triangles we now have, so the aim is to get as close to 50% of the remaining two triangles as possible?

  • $\begingroup$ The problem with your suggestion is that Alice can choose to cut through the smallest two of the three triangles, and leave the largest for herself. $\endgroup$ – Joe Z. Feb 21 '13 at 16:25
  • $\begingroup$ Also, is the triangle equilateral? $\endgroup$ – Joe Z. Feb 21 '13 at 16:26
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    $\begingroup$ The problem is "affine invariant". Therefore assuming the triangle to be equilateral helps the intuition, but is no restriction of generality. $\endgroup$ – Christian Blatter Feb 21 '13 at 16:36
  • $\begingroup$ So it's fair to assume that the point should be placed as centrally as possible? But if the triangle is not equilateral, how does he determine this point and what percentage can he expect to get? $\endgroup$ – user61067 Feb 21 '13 at 16:39

You can consider the triangle to be equilateral, as you can make any triangle equilateral with a linear transformation. That transformation will preserve the ratio of areas. The centroid is the obvious point to pick. If Alice cuts parallel to a side, she leaves Bob $\frac 49$ because the centroid is $\frac 23$ of the way along the altitude.

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  • $\begingroup$ I think I'm going to have to do a bit of reading on centroids before fully understanding this, but thank you :) $\endgroup$ – user61067 Feb 21 '13 at 16:53
  • $\begingroup$ More succinct than my answer (and more accurate explanation), I suppose. +1. $\endgroup$ – Joe Z. Feb 21 '13 at 16:55
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    $\begingroup$ @user61067: the centroid is the intersection of the medians, the lines from each angle to the midpoint of the opposite side. It is the center of gravity of the triangle. Wikipedia has a short discussion. $\endgroup$ – Ross Millikan Feb 21 '13 at 16:57
  • $\begingroup$ Ah I see, thanks again, you too Joe! $\endgroup$ – user61067 Feb 21 '13 at 17:52
  • $\begingroup$ @RossMillikan how would I prove that if she was to cut through that point using a line that isn't parallel to one of the sides..she would end up with less pie ? $\endgroup$ – user61067 Feb 27 '13 at 17:35

If the triangle is equilateral, it seems like the best answer would be for Bob to pick the center of the triangle, in which case Alice cuts one of the corners off and leaves Bob with 4/9 of the cake.

The reason that any point other than the center will not work is because Alice can simply cut parallel to the furthest side from the point, and will leave Bob with less than 4/9.

And if the problem is affine-invariant like Christian Blatter claims it to be, then the point on non-equilateral triangles should be a specific center called the centroid, the intersection of the three medians. And the percentage Bob can expect to get is still 4/9 as usual.

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  • $\begingroup$ But what if it isn't equilateral? $\endgroup$ – user61067 Feb 21 '13 at 16:37
  • $\begingroup$ @user61067 See the edit. $\endgroup$ – Joe Z. Feb 21 '13 at 16:44
  • $\begingroup$ It is not $25\%$ because the centroid is not half way along the altitude. $\endgroup$ – Ross Millikan Feb 21 '13 at 16:51
  • $\begingroup$ Oh wait, you're right. Whoooops. $\endgroup$ – Joe Z. Feb 21 '13 at 16:54
  • $\begingroup$ Not sure how I missed that. $\endgroup$ – Joe Z. Feb 21 '13 at 16:55

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