- 08 Jan '08 19:17Actually only one pancake. Given one finite area, two-dimensional pancake, prove that there exists two perpendicular straight lines that cut the area of the pancake into four equal pieces. i.e. prove that you can cut a pancake into four equal pieces with two knife strokes, perpendicular to each other.

Genaralize to n dimensions...

If the pancake is a 3-4-5 triangle, how would you cut it? - 09 Jan '08 18:43 / 2 edits

Isn't this just a circle with an X drawn on it through its midpoint?*Originally posted by David113***Actually only one pancake. Given one finite area, two-dimensional pancake, prove that there exists two perpendicular straight lines that cut the area of the pancake into four equal pieces. i.e. prove that you can cut a pancake into four equal pieces with two knife strokes, perpendicular to each other.**

- 09 Jan '08 19:42 / 2 edits

OK, I think I have something.*Originally posted by David113***Actually only one pancake. Given one finite area, two-dimensional pancake, prove that there exists two perpendicular straight lines that cut the area of the pancake into four equal pieces. i.e. prove that you can cut a pancake into four equal pieces with two knife strokes, perpendicular to each other.**

Genaralize to n dimensions...

If the pancake is a 3-4-5 triangle, how would you cut it?

Building on results from the original thread, we know an infinite number of centre-lines exists that will cut a pancake in half. This is because when sweeping out the area of a pancake using a straight line, the area swept is continuous and ranges from 0 to 1. Since it is continuous, it must take on all values between 0 and 1, including 0.5 which gives us the existence of at least one centre-line. Since this derivation works irrespective of the orientation of the pancake, it is possible to rotate the pancake through any given angle and obtain the same result. Therefore, there are an infinite number of centre-lines.

Now, consider a pancake with an arbitrary centre-line already drawn on it. On this pancake, we will draw another centre-line (not necessarily perpendicular to the first). We will name the newly created regions counterclockwise as follows:

B | A

--|---

C | D

Since the two lines we have drawn are centre-lines, it follows that, in terms of area:

A + B = C + D

and:

B + C = A + D

Adding these equations, we get:

A + 2B + C = A + C + 2D

2B = 2D

B = D

Subbing this result in to the first equation gives us:

A + (D) = C + D

A = C

Since A = C, and B = D, if we choose an appropriate pair of perpendicular centre-lines such that A = B, we will have shown that A = B = C = D = (A+B+C+D)/4 as required.

Let's go back to our original construction, with one arbitrary centre-line drawn on the pancake. We will now draw another centre-line in a special way. To begin, we will have the second centre-line coincide with the first. In this case, region A has an area of 0 and region B has an area of half the total area, or (A+B+C+D)/2. Also, the angle of intersection of the two centre-lines is 0 radians. By sweeping out the angle of intersection between the two centre-lines (whose*point*of intersection may move) from 0 to pi radians, the area of region A will be swept out from 0 to half the total area. Using the result from the first paragraph, since this area function ranges continuously between 0 and (A+B+C+D)/2, there must be at least one point where A = (A+B+C+D)/4 as required.

Now, this point may not coincide with an angle of pi/2, which is also required. However, it is possible to rotate the first centre-line incrementally (i.e. d(theta)) in the same direction and repeat the process. This will result in a graph that is shifted left by d(theta) units and down by the value of A at d(theta) on the original graph. Of course, the derivation above does not depend on the orientation of the original centre-line since it was arbitrarily chosen, so the existence of the A = (A+B+C+D)/4 is still guaranteed but now it will occur at a different angle of intersection.

Now, it is not possible for the point A = (A+B+C+D)/4 to always lie either to the left or the right of pi/2 on the graph for all starting centre-lines. To see this is so, consider the graph of area B, which is just the graph of area A reflected in the line theta = pi/2. If the point lies to the left of pi/2 on the graph of A, it must lie to the right of pi/2 on the graph of B. If we move the first centre-line to coincide with our second centre-line and begin sweeping out the area, our new graph of A is just the mirror reflection of the original graph of B! Since the angle dividing the pancake must occur on either side of pi/2, and since the location of the angle ranges continuously between at least these two point due to the incrementally changing nature of our construction, it is therefore**guaranteed**that for some initial centre-line, at least one perpendicular centre-line can be drawn that divides the pancake into 4 equal pieces.

Whew!! Sorry if that's confusing, it looks so much simpler on paper...anybody got a pen!? - 10 Jan '08 00:00

If already have your proof on paper just upload it here http://tinypic.com/ cause it is easyer on the eyes with true math notation and the drawings come better too.*Originally posted by PBE6***OK, I think I have something.**

Building on results from the original thread, we know an infinite number of centre-lines exists that will cut a pancake in half. This is because when sweeping out the area of a pancake using a straight line, the area swept is continuous and ranges from 0 to 1. Since it is continuous, it must take on all values between 0 and 1, ...[text shortened]...

Whew!! Sorry if that's confusing, it looks so much simpler on paper...anybody got a pen!?

It could be cool if we had some math fonts to play with. - 10 Jan '08 03:04 / 2 edits

i think i follow you... an easier way to put it might be: given an initial center-line*Originally posted by PBE6***OK, I think I have something.**

Building on results from the original thread, we know an infinite number of centre-lines exists that will cut a pancake in half. This is because when sweeping out the area of a pancake using a straight line, the area swept is continuous and ranges from 0 to 1. Since it is continuous, it must take on all values between 0 and 1, ...[text shortened]...

Whew!! Sorry if that's confusing, it looks so much simpler on paper...anybody got a pen!?**L**, there exists a line**M**at some angle theta with respect to that center-line, that simultaneously bisects the other two regions. that proof is solid, i won't repeat it.

now, consider a few cases: if**L**and**M**are perpendicular, we're done.

if theta < 90, consider the change in theta as line**L**is rotated an angle phi (and**M**continues to be defined as the bisector of the other regions). all of these rotations will be continuous functions (not sure if this is true?) and when phi = 180, theta is now clearly > 90 - in fact it should be the supplement of theta at the beginning when phi = 0!

Thus theta = 90 for some value of phi between 0 and 180, so a pair of these perpendicular bisectors must exist. i hope this is valid, because my brain hurts.. but my intuition says that a continuous rotation of phi wouldn't necessarily relate to a continuous change in theta. and i'm either not savvy enough, or just entirely disinclined to try and thoroughly nail that down - 11 Jan '08 13:25

How about this - in the case of a convex pancake.*Originally posted by David113***The proof seems OK to me...**

*** here is another problem: ***

Given one finite area, two-dimensional pancake, prove that there exists a straight line that cuts the area of the pancake AND ALSO ITS BOUNDARY into two equal pieces.

We've already established that, given a pancake and a point outside the pancake, we can draw a line through the point that halves the pancake.

So, draw a circle enclosing the pancake. Choose a point A and draw such a line through it. This will meet the circle again at a point B. Traverse one arc of the circle from A to B, at each point drawing a line that halves the pancake.

The line will also split the circumference. Consider the length of the right-hand side (from the point of view of the initial point) as a proportion of the total. Either this is 0.5 (job done), less that 0.5, or greater than 0.5.

Let's say it's less that 0.5. When we reach B, the line is the same line, but now the proportion will be greater than 0.5. Since this proportion will change continuously as we traverse the arc [THIS IS THE BIT THAT MAY NEED FURTHER WORK], and it has gone from < 0.5 to > 0.5, there must be a point where it is equal to 0.5. QED.

I've realised that if the pancake is not convex, this and the other problems run into trouble. Consider a spiral shaped pancake. Any line that splits it into areas of equal size will be splitting the pancake (and its circumference) into*more*that two pieces. - 13 Jan '08 03:57your spiral pancake comment gets at what my intuition was saying about how a rotation of an initial bisecting ray would not necessarily be continuous with regard to the rotation of an intersecting bisector: i.e. with a concave pancake, a rotation of your reference bisector could lead to a wild change in the angle at which it meets the line that simultaneously bisects the two new regions. with a convex pancake, it is immediately apparent that we'd be ok. the original pancake theorem allows for any shaped region (i believe the only constraint is a finite-length boundary)

we may be overlooking something? or maybe we're onto something lol - 14 Jan '08 14:15I'm not sure if this helps anything at all, but is it possible to transform an arbitrary pancake (even possibly a concave one) into a topological equivalent (like a circle) on a different plane, prove the result is true for that circle, and then find the transformative inverse to bring the result back to pancake land?

I don't know much at all about topological calculations, but in "The Elegant Universe" which I'm reading now, physicists used this approach to solve some very difficult problems having to do with curled-up spatial dimensions. Can anyone illuminate the usefulness or uselessness of this approach? - 14 Jan '08 14:19

It might be useful in some cases. But the problem I can see here for the general case is that once your pancake has been transformed into a nice circle, your straight lines are anything but! So I think you've probably just exchanged one difficulty for another one just as bad.*Originally posted by PBE6***I'm not sure if this helps anything at all, but is it possible to transform an arbitrary pancake (even possibly a concave one) into a topological equivalent (like a circle) on a different plane, prove the result is true for that circle, and then find the transformative inverse to bring the result back to pancake land?**

[Not a definitive answer - just what my intuition is telling me] - 14 Jan '08 18:55 / 1 edit

I have the opposite intuition about this. Finding the area of part of a circle bounded by a squiggly line is a fairly easy calculation to do, even if we have to resort to a piece-wise integration. However, the question does not require us to determine the actual method of finding the line that cuts the pancake and its boundary in half - we are only required to prove or disprove its existence.*Originally posted by mtthw***It might be useful in some cases. But the problem I can see here for the general case is that once your pancake has been transformed into a nice circle, your straight lines are anything but! So I think you've probably just exchanged one difficulty for another one just as bad.**

[Not a definitive answer - just what my intuition is telling me]

Does anyone know a general formula for transforming a convex shape described by a parameterized function (t,**f(**t**)**) into another parameterized function (s,**g(**s**)**)?

EDIT: A more important question - can any result determined in this new space or shape be carried back to the original shape/space? Or do very specific criteria need to met?