Treebeard's Stumper Answer

Christmas CutUps
You can make a perfect fivepointed star from a circle of paper with just one straight cut if you fold it right first. But how? You can cut a holiday pie into two pieces with one straight vertical cut, four pieces with two cuts, or seven pieces with three cuts, if you're careful and don't move the pieces between the cuts. How many pieces can you make with ten cuts? Is there a general formula for any number of cuts? I suppose you can experiment with lines on paper, but it will be much more satisfying to experiment with real holiday pies!
0 cuts, 1 piece. 1 cut, 2 pieces. 2 cuts, 4 pieces. 3 cuts, 7 pieces. I don't know if it's actually possible to draw all the pieces of a pie after ten cuts, but I'll have to try! A better strategy for this stumper is to think about the pattern and understand the effect of making one more cut. This stumper can be extended in many ways:
 Is it possible to draw the pie after 410+ cuts with pleasing symmetry like I started doing above?
 With three different (noncolinear) cuts I get a maximum of 7 pieces and a minimum of 4 pieces if the cuts don't cross. What is the minimum number of pie pieces for additional cuts? Is every inbetween value always possible?
 What about slicing a solid cheese ball into pieces if the straight cuts can come from any direction? This is very hard to visualize, but the math is similar.
 Cutting a circle is the same as cutting a square, since both plane shapes are convex. But what about cutting a concave shape like a crescent moon or a star? What about slicing a banana?
 The big question on my mind is how many equal area/volume pieces are possible with n cuts. In my picture of the pie with three cuts, there are seven unequal pieces. It's easy to get the minimum of six equalarea pieces if all cuts meet in the center like a sliced pizza. Is it also possible to adjust the cuts to give seven equal area (not shape) pieces? Is this general?
 I started this by swiping a nice picture of a bakery pumpkin pie, but after I reduced it to 16 colors, it looks more like the end of a pine log. Now I realize there's another practical stumper here about how to best cut 2x4s (etc.) from a whole log. I'll save that for a future stumper! I love the way one question unexpectedly leads to another. Good questions take us farther than answers!
3 cuts, 6 equal pieces.
You can cut a pie into 56 pieces with ten straight cuts if you're careful. The first cut divides the whole pie into two. The second cut across the first adds two more pieces. The third cut across both lines will add three more pieces. So the nth cut will cross n1 lines and add n new pieces. After ten cuts, the sum is 1 + 2 + 3 + ... + 10 plus one more for the whole pie we started with. For n cuts, that's [n (n+1) / 2] + 1. Keep reading for my best attempt at drawing all ten slices, as well as the secret Betsy Ross method of making a star with a single cut!
A mandala of ten slices and 56 pieces!
Can you color it with just four colors?Notes:
Several DMS students figured the answer once they noticed the rule and the pattern. You can only get the maximum number of cuts if you follow the rule that every new cut must cross every other cut and there can be no intersections of more than two lines at once. Graybear sent this clear explanation of the pattern:
(# of cuts, max # of pieces):
(0,1); (1,2); (2,4); (3,7); (4,11); (5,16); (6,22); (7,29); (8,37); (9,46);(10,56); ...
(n, n^2/2 + n/2 + 1).Reason: each successive cut 'n' can cross 'n1' previous cuts; therefore it slices through 'n' separate pieces and creates 'n' more. I.e. the first cut creates one more piece; the second cut creates two more pieces; the third cut creates three more pieces; etc. The general formula is similar to Gauss's famous equation:
n(n1) Sum of # (1n) =  2except we have to add 1 for the original piece, so it becomes:
n(n+1) n^{2} n  + 1 or  +  + 1 2 2 2Drawing the ten cuts was more difficult, but I found a pattern there too. Start with two lines crossing and cross them both with a third line to make the classic anarchy symbol (left). Then replace that line with two new "butterfly" lines that cross in the center below the others for a total of four lines (right). Move the intersection up if you can to make way for later lines.
3 cuts, 4+3 = 7 pieces 4 cuts, 7+4 = 11 pieces Now do it again. Cross all the lines with a single cut (left), and replace that line with two more that cross in the center below the others (right).
5 cuts, 11+5 = 16 pieces 6 cuts, 16+6 = 22 pieces With some fiddling, it should be possible to show any number of cuts, though the pieces get pretty small. I'm sure it's not possible to get equal areas, but it is probably possible to get more equal areas than I managed to. I decided to quit while I was ahead! (And where do equal areas fail? It is possible with two cuts of course, and it looks possible with 3 cuts into 7 equal pieces. But it's hard to get even visible pieces with ten cuts!))
Graybear corrected me on the minimum number of pieces, d'oh!
I get a minimum of 4 pieces by not crossing any previous cuts (picture a circumscribed triangle). The minimum number of pieces after 'n' cuts is 'n+1' pieces. Is every inbetween value always possible? Yes, by adjusting how many previous cuts you choose to intersect.Graybear also figured out a way to visualize slicing a solid cheese ball into pieces. This is very hard to visualize, but the math is similar:
I DEFINITELY agree with both parts of the last sentence, but I finally figured out how to visualize it (without paper, I might add) and the math is very similar, actually building on the first part of the problem. Imagine, instead of a pie, you have a sphere. After the first cut, picture a crosssection of the sphere that looks like the pie after one cut showing the two pieces. Your section cuts through both pieces creating two more for a total of four. Imagine the next crosssection that looks like the pie after two cuts showing four pieces. This section then cuts through all four creating four more for a total of eight. Adjust your next crosssection so that the three previous cutting planes can appear as the pie after three cuts so you can create seven more pieces. Continue on so the results become: (0,1); (1,2); (2,4); (3,8); (4,15); (5,26); (6,42); (7,64); (8,93); (9,130); (10,176); (n, n^3/6 + 5n/6 + 1).That general answer can also be written as(n1)n(n+1) n^{3} 5n  + n + 1 or  +  + 1 6 6 6Thanks Graybear, I had that answer, but I couldn't visualize it. I'm sure crescent moons and bananas and stars produce similar polynomial solutions, but those are stumpers for another day!
I almost forgot my other stumper. The story is that George Washington's original sketch for the American flag had 6pointed stars. But seamstress Betsy Ross preferred a 5pointed star. (Guess why!) When the committee protested that it was too hard to make that shape, she took a piece of paper, quickly folded it, and produced a perfect fivepointed star with a single snip. The committee was so impressed that they agreed with her suggestion. Here's how she did it:
Fold a circle of paper in half and divide it into five equal pieces as shown by the blue lines. Fold along the lines as shown in the middle picture, and cut along the red dotted line. By changing the angle of the cut, you can make different shaped stars. There are stepbystep instructions for making the folds at the Betsy Ross Homepage. (But good luck explaining it to a group of 66 kids!)
Here are some links for further research:
 These are two classic problems (#161 and 164) from puzzle master H.E Dudeney's Amusements in Mathematics (Dover, 1958). The other puzzle master Sam Lloyd posed some variations (#101 and 112) in Mathematical Puzzles of Sam Lloyd (Dover, 1959)
 W.W. Sawyer uses the pie puzzle as an exercize (without answer) in his book The Search for Pattern (Penguin, 1970, unfortunately out of print). Sawyer explains the easy way for finding a quadratic formula for a number series by considering successive differences:
n: 0 1 2 3 4 5 6 7 f(n): 1 2 4 7 11 16 22 29 dif1: 1 2 3 4 5 6 7 (difference between successive f(n)) dif2: 1 1 1 1 1 1 (difference between successive dif1)The second differences are all equal, so we know it's quadratic. Look at the first numbers (in bold):Therefore the equation is 1/2 n^{2} + 1/2 n +1, which is the answer.
 dif2 is always 1, so take 1/2 of that for the x^{2} coefficient.
 dif1 minus 1/2 of dif2 is 1  1/2 = 1/2. That's the x coefficient.
 f(n) for n=0 is 1, and that's the constant.
This method always works when the second difference is constant. The method can be generalized for higher powers (and higher differences), though the details get tricky. I'm surprised I can't find a good treatment on the Web. A cubic example is discussed here.
 David Molnar has a page on Cutting Fruit with even more cutup questions. There's also a classroom worksheet if you need it.
 You can also cutup words and pictures and music into interesting travesties of the original. Postmodern art and hiphop music are based on remixing other sources. William S. Burroughs used his cutup method to reassemble works by Shakespeare and Rimbaud. That sounds like "conceptual art", but he once did an entire cutup Time magazine that I actually saw for sale at City Lights Bookstore in San Francisco back in the 60s. (I didn't buy it, d'oh! It's a collector's item now.) If you cut a text into single letters and reassemble it, you can make your own universal library! See my stumper The Universal Library of Stumpers (23 March 2001) and the discussion by Quine. There's a cutup/travesty maker you can play with on the Web.
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