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Let's say you have a donut. You are allowed to slice it 3 times. Each slice must be a perfectly straight cut. What is the highest number of donut pieces you can end up with after 3 slices?

Assume that no crumbs are created during the slicing.

Also assume that no pieces move until after you have finished all 3 slices. That way while you are making your third slice, the pieces made from the first and second slices don't start moving and falling off.

Please hide your answers with the spoiler markup!

pacoverflow
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6 Answers6

18

You can cut it into 13 pieces.

I couldn't draw the picture, but I found this website that has already drawn it for me:

Reference: http://www.hunkinsexperiments.com/pages/doughnuts.htm

Unfortunately I cannot include the picture here as it is copyright.

pacoverflow
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Kenshin
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    This is effectively a link only answer and can only remain so due to copyright. It should really be a comment. – Shoe Nov 08 '14 at 12:56
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    @Jefffrey, the answer to the question is simply 13. I thought the link assisted in seeing how it is done, but if you prefer I can describe the picture in words? – Kenshin Nov 08 '14 at 13:11
  • That's also an option. – Shoe Nov 08 '14 at 13:17
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    I saw this question in The Second Scientific American Book of Mathematical Puzzles and Diversions On page 149 is a formula for the maximal number of pieces obtainable with n cuts, $\frac{n^3 + 3n^2 + 8n}{6}$. (It has the same picture, but with better quality and probably with expired copyright) – DenDenDo Nov 08 '14 at 16:57
  • @DenDenDo, ty bobson – Kenshin Nov 08 '14 at 17:48
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    Good link, DenDenDo. Mew, I think you should quote that one paragraph with the formula on page 149 as well as include the image from page 150 (available here: http://www.reanimationlibrary.org/catalog/system/scans/1761/large/100912_0150.jpg). That should be considered "fair use" under copyright law, as long as you mention you got it from the book. – pacoverflow Nov 08 '14 at 19:01
  • This is better than the solution (12 pieces) found in Steven Krantz, Techniques of problem solving, solution 2.39, p. 391. – Watson Aug 23 '16 at 10:23
  • Is this really the maximum? From the diagram the top has 5 small pieces, but the bottom has 6 small pieces. My gut feeling is that we can get another small piece at the top too, bringing the total to 14. – Dmitry Kamenetsky Oct 02 '20 at 02:20
9

You can create 10 pieces. Make two cuts that are perpendicular to the table as shown, tangent to the hole, creating five pieces. Then, slice the donut parallel to the table, splitting each piece into two.

enter image description here

pacoverflow
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xnor
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4

Edit: This works if we're talking about a filled donut. For toroidal donuts better answers are given already.

You can get

8 If you make each cut intersect all of the others, for example by making the all perpendicular to each other.

This is provably the maximum unless you get tricky and distort the shape:

Each cut can, at best, split each existing piece in two, doubling the current number. We start with 1 piece, 1 cut gives 2, 2 cuts gives 4, 3 cuts gives 8.

frodoskywalker
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enter image description here


My answer is 9 as it can be seen in the picture. (without using 3D technique) :)

Nai
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I can create

8 pieces by cutting along the axis:

With cuts  

1: Horizontally through the middle of the piece.
2 and 3: Like a cross vertically.

pacoverflow
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v010dya
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Assuming we're cutting just perpendicular to the surface, the maximum is

9 Pieces. This is by first cutting at bearing 000, to the left of the centre but still going through the circular gap in the centre, Then, rotate the donut by 120 degrees and make the same cut. Repeat this once more and you should have 9 separate pieces if the distance from the centre that you used was correct. Simply scale the distance from the centre for each cut until the cut clips the hole, but also crosses with another cut before doing so. Of course, if we allow non-perpendicular cuts to the surface...

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