math and pancakes

Originally posted by adam warlock
Doesn't the same type of argument also applies here? We can 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?

sweeping both pancakes at the same time would cut the TOTAL area in half, but likely wouldn't cut EACH pancake in half... or possibly you mean to sweep them each from the same point in which case it's very unlikely that the two bisectors would be the same line

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]... work.

Oh, and as far as I can tell this doesn't require the pancakes to be non-overlapping.

it does need a little work to be more rigorous, as you said, but this is some nice cute reasoning. you definitely got the spirit of the proof here with taking the bisectors, and their angles from a place of reference, and then looking at a sweep where the angles between the bisectors goes from negative to positive. it may not be entirely accurate, (in very extreme cases) that as you move further up your vertical line the angle of the bisector changes continuously. though your point of reference is much easier to describe if it indeed holds in more extreme cases ... let the incident angle with blob A's bisector = a and let the angle with blob B = b, and then look at b - a ? definitely looks good for most cases, but i haven't thought about the "weird" cases enough to take it as gospel. nice work.

Originally posted by Aetherael
definitely looks good for most cases, but i haven't thought about the "weird" cases enough to take it as gospel. nice work.

Fortunately you ruled out the really weird cases by stating they had to be convex pancakes 🙂

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]... work.

Oh, and as far as I can tell this doesn't require the pancakes to be non-overlapping.

There are an infinite number of lines that will cut a pancake in half, because the proof and construction method outlined previously does not depend on orientation. Since this is the case, you can construct a centre-line through any point on the edge of the pancake.

My solution is as follows:

Any two pancakes can be joined by a straight line that passes somewhere through the body of both pancakes. Let's take a straight line that coincides with the centre-line for one of the pancakes, our "control" pancake. This line will always cut one pancake in half, and will pass through the other in some fashion.

Wherever the line passes through the second pancake, draw two new centre-lines through both intersection points. These new centre-lines will intersect somewhere in the body of the second pancake. Now, if the point of intersection of the new centre-lines lies to the right of the "control" centre-line, rotate the whole system of pancakes to the right, or vice versa. Draw a new centre-line through the control pancake and repeat the process until the new centre-lines intersect with the "control" centre-line. Since the "control" centre-line cuts the first pancake in half, and coincides with the centre-line that cuts the second pancake in half, you have constructed a line that cuts both pancakes in half.

Originally posted by mtthw
Fortunately you ruled out the really weird cases by stating they had to be convex pancakes 🙂

you're totally right. 🙂 nice work - the proof for allowing concave blobs is kind of cute too: involves taking an arbitrary circle that contains both "pancakes" and doing a similar area sweep (the setup is a little more involved, and not quite as elegant i.m.o as your point moving along a straight line). However, it allows for more generality and shows you can cut funny shaped blobs in half too.

for anyone interested in the multidimensional equivalent of the "pancake theorem", check out the "ham sandwich theorem." it sounds like a joke, but i'm not kidding.

check it out: http://en.wikipedia.org/wiki/Ham_sandwich_theorem

Originally posted by PBE6
There are an infinite number of lines that will cut a pancake in half, because the proof and construction method outlined previously does not depend on orientation. Since this is the case, you can construct a centre-line through any point on the edge of the pancake.

My solution is as follows:

Any two pancakes can be joined by a straight line that passes ...[text shortened]... he second pancake in half, you have constructed a line that cuts both pancakes in half.

i'm not sure i follow... but isn't it entirely possible that the first "center line" of the control pancake (first, assuming you choose one the DOES interesect the other pancake) will cut the second pancake in such a way that the intersections provide new center lines that don't intersect the control pancake? imagine the creation of two points whose center lines end up being almost perpendicular to the original center line?

with this system can we set up a "sweep" type scenario that continuously moves from "too far left" to "too far right" and thus must go through "just right"? 🙂 or maybe, as a consolation prize, an iterative process that necessarily gets closer and closer to correct through each iteration (i.e. in the limit, we achieve the sought result)?

i think your approach has a lot of merit, but unless i'm understanding it incorrectly, i think it needs to be more thorough. good luck!