Yea, they were taking it into 3 dimensions which seems to refer to a similar puzzle of cutting a round cake with only 3 cuts into 8 pieces of equal volume.
Two vertical cuts quatering the cake ’ + ’ and then a horizontal cut through the cake
Let’s say the “linearity” of a particular configuration of cuts is the maximum number of pieces a new line could cut across. Notice that if adding a cut to a configuration gives it linearity N (i.e., after the new cut, there’s some way to lay a line across it cutting across N pieces), then at most one of the relevant piece-separators is due to the new cut, so prior to the new cut, the linearity must have been at least N - 1. I.e., each cut increases the linearity by at most 1. Also, the pizza clearly starts out with linearity 1. Thus, before the Nth cut, the linearity is <= N.
Furthermore, the number of new pieces a cut creates (by splitting old pieces) is clearly at most the linearity of the configuration prior to the cut. Combining this with the previous observation, we find that the Nth cut creates <= N new pieces. In other words, after N cuts, a total of <= 1 + 2 + … + N = N * (N + 1)/2 many pieces is created. Adding the 1 piece the pizza starts with, we see that after N cuts, the pizza is left in <= 1 + N * (N + 1)/2 many pieces.
In the particular case where N = 4, this gives us an upper bound of 1 + 4 * 5/2 = 11 many pieces.
I can get 16 congruent pieces with 4 cuts. First, cut along the diameter. Then, take one half and stack it on top of the other half, and make the second cut along the midline. For the third and fourth cuts, stack and cut on the midline again.
Also clearly the maximum possible, for the problem allowing stacking but no folding. 
Again, you’re making a 2D exercise into a 3D one. There’s no such thing as “stacking” things on top of each other in a two dimensional plane.
But you got cheese on the bottom of the other pieces. That is a fail.
Ah, but what if the pizza comes with an unlimited number of those little plastic picnic tables that go in the middle to keep the box of the cheese, huh? WHAT THEN??!
You flip one piece over so that the cheese side faces the cheese side. After that crust is on the outside and stacking is no longer an issue.
Ha ha. If you can cut this piece of paper into four equal sections, I’ll give you a quarter.
Let me try to run post #22 through the degeekulator.
Without relying on a visual I, I think the best way to explain it verbally is:
The first cut (1) can only hope to produce two sections, a and b. Hopefully that doesn’t need its own proof.
The second cut (2) can be made in one of two ways.
[ul]
[li]Either entirely within section a or section b, the result is identical to that of cut #1 (splitting a single section into two sections). This results in a single additional section c, for a total of three sections.[/li][li]Make a cut that intersects cut #1. This results in two additional sections, for a total of four sections.[/li][/ul]
From this last step we observe that, if the choice is either intersecting the previous cut or not intersecting it, the choice that yields more sections is intersecting the previous cut. Let’s apply this newly acquired practical knowledge when making cut #3 
The third cut (3) can be made in one of four ways:
[ul]
[li]Entirely within any single section (one additional section produced)[/li][li]Intersecting a single previous cut (two additional sections produced)[/li][li]Intersecting the previous two cuts’ intersection point (two additional sections produced)[/li][li]Intersecting both previous cuts (splitting three existing sections in two, and resulting in three additional sections)[/li][/ul]
We can see that the fourth option yields the largest number of sections.
The fourth and final cut should be made in such a way that it intersects all three previous cuts. If we continue on with additional cuts, and always wish to generate the maximum number of sections, we must always cut in such a way that we intersect all previous cuts.
Wait, wait - what if the four lines represent the four blades of a blender? You’d approach an infinite number of slices then!
I don’t know why I’m screwing with this thread. I’ll stop now.
Martin Gardner died just last Saturday.
RIP
http://www.nytimes.com/2010/05/24/us/24gardner.html?ref=obituaries
Oh, man, he was a great popularizer. Are there any compendia of his work?
You can always check Amazon.com, for instance, to see what’s available new and used.
Over the last few years he was still at work writing updates for new editions of his Mathematical Games columns. They’re worth a gander if you don’t have any of his published works already.
Now the next problem is: what orientation of these four cuts give pieces with the smallest difference in area? And what statistical measure of difference is the best to use?
where are the geeks of this board ?
My gut feeling is you could make the pieces identical in area so it would not matter what measure you used. If this is not possible, there is no correct answer for the “best measure.” Three obvious choices are: max area - min area, variance (or std dev), and sum of the 55 absolute differences.
I don’t think it’s possible to make them identical in area. Looking at the figure and attempting to move the cuts mentally, it seems to me that moving the lines to shrink one of the larger pieces inevitably affects at least one of the smaller pieces negatively.