- 07 Jan '08 15:13

imagine you have two very different looking, very different sized pancakes... now you are looking for a single straight cut that cuts EACH pancake exactly in half (in terms of area): not one cut that cuts the total area in half. it's not entirely obvious how to find where to cut or even necessarily that the cut exists - any clearer?*Originally posted by coquette***How could there not be?** - 07 Jan '08 15:30

Do you want an existence theorem or an operational theorem that tells you how to do it?*Originally posted by Aetherael***imagine you have two very different looking, very different sized pancakes... now you are looking for a single straight cut that cuts EACH pancake exactly in half (in terms of area): not one cut that cuts the total area in half. it's not entirely obvious how to find where to cut or even necessarily that the cut exists - any clearer?**

The existence theorem is easy the operational one would take a lot of time - 07 Jan '08 16:00

unless i'm overlooking an easy way to prove the existence without an operational method, then i'm pretty sure the two go hand in hand... my method involves invoking the mean value theorem, if that helps*Originally posted by adam warlock***Do you want an existence theorem or an operational theorem that tells you how to do it?**

The existence theorem is easy the operational one would take a lot of time - 07 Jan '08 16:08

Ok: here is my*Originally posted by Aetherael***unless i'm overlooking an easy way to prove the existence without an operational method, then i'm pretty sure the two go hand in hand... my method involves invoking the mean value theorem, if that helps***proof*:

You got the pancake. Let's think of a straight line that is outside the pancake. This straight serve us to indicate what portion of the pancake is being covered with the straight line. We start from left and go to right. When we are athe left and aren't touching the pancake the are is 0% as we go to right the area of the pancake that's being covered increases. It reaches a point that we have covered all the pancake and if we move more to the right the covered area is still 100%. Since the area covered is obviously a continuous function there must be a point were we have covered 50% of the pancake.

Sorry for the written text but without math fonts functional notation would look too funny. And I hope I got my idea across too. - 07 Jan '08 16:11

this is exactly right for a*Originally posted by adam warlock***Ok: here is my***proof*:

You got the pancake. Let's think of a straight line that is outside the pancake. This straight serve us to indicate what portion of the pancake is being covered with the straight line. We start from left and go to right. When we are athe left and aren't touching the pancake the are is 0% as we go to right the area of the pa ...[text shortened]... out math fonts functional notation would look too funny. And I hope I got my idea across too.*single*pancake... but now we must extend this somehow to allow for bisecting two separate pancakes all with a single cut - nice job with the first piece of the proof though; you've found a way to cut one of the pancakes in half, so that's a start - 07 Jan '08 16:18 / 1 edit

let me rephrase... these pancakes are not by definition circular. in fact they could be very oddly shaped blobs. choosing them to be convex was a way of trying to avoid arguments about discontinuous blobs, but that may have been misleading insofar as making this seem simpler and more obvious than it is.*Originally posted by doodinthemood***Any line which goes through the centre of a pancake cuts it in half. Thus, any line that cuts through the centres of both pancakes cuts both of them. In any situation, it is possible to get a straight line between two points so it is always possible.**

imagine you've got a couple of weird shaped blobs on a plate, but no holes in the middle of the blobs. prove that you can cut them both in half with a single cut of a big ol' knife (you can even bring out your katana, if you so choose).

note: with oddly shaped figures, you could get vastly different results in terms of area splitting, by cutting though a particular point in the figure at a different angle. i.e. a straight up and down cut through the centroid could be way different from a horizontal cut or a slight angle or a 45, etc. - 07 Jan '08 16:21 / 1 edit

lolol...haven't read your original post with enough attention... I have a habbit for doing it. I'll try and think about the whole proof now.*Originally posted by Aetherael***this is exactly right for a***single*pancake... but now we must extend this somehow to allow for bisecting two separate pancakes all with a single cut - nice job with the first piece of the proof though; you've found a way to cut one of the pancakes in half, so that's a start

Edit: By the way - the pancakes don't overlap do they? - 07 Jan '08 16:26

in my version of the problem they do not.. but a case possibly could be made for stacking the pancakes without loss of generality? provided that orientation of the pancakes is preserved? that's a tough call, so i will say that the pancakes do not overlap unless you show me they can be stacked wlog. good luck with the proof!*Originally posted by adam warlock***By the way - the pancakes don't overlap do they?** - 07 Jan '08 16:33

Doesn't the same type of argument also applies here? We can*Originally posted by Aetherael***in my version of the problem they do not.. but a case possibly could be made for stacking the pancakes without loss of generality? provided that orientation of the pancakes is preserved? that's a tough call, so i will say that the pancakes do not overlap unless you show me they can be stacked wlog. good luck with the proof!***sweep*both pancakes at the same time and have 0% and 100% two pancake area. So in between there must be a point that goes to 50%...

Very unsavoury solution but a solution nonetheless? - 07 Jan '08 16:49 / 2 editsMy sketch proof goes something like this:

Arrange the pancakes roughly horizontal, and draw a vertical line to the left of them (not overlapping either one).

For each point on that line, using a similar argument to Adam's above, there must be a unique line that halves each pancake.

Now consider the angle between these lines. From one end of the vertical line to the other, this goes from positive to negative. So (by the MVT) there must be a point where this angle is zero. At this point the two lines are the same.

Any good? That probably needs some work to make it rigorous, but it seems to me that the approach should work.

Oh, and as far as I can tell this doesn't require the pancakes to be non-overlapping. - 07 Jan '08 16:51

Seems to be good enough to me.*Originally posted by mtthw***My sketch proof goes something like this:**

Arrange the pancakes roughly horizontal, and draw a vertical line to the left of them (not overlapping either one).

For each point on that line, using a similar argument to Adam's above, there must be a unique line that halves each pancake.

Now consider the angle between these lines. From one end of the verti ...[text shortened]... probably needs some work to make it rigorous, but it seems to me that the approach should work.