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?
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, 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
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!?