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Can you cut a pizza (circle) into 12 congruent pieces, such that half of them have crust (circle boundary), while the other half do not? The pieces must have the same shape and area, but can be mirrors of each other.

Bonus: Can you do it with identical pieces, where pieces are not mirrors of each other?

Dmitry Kamenetsky
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    For the record I like my pizza with crust :) – Dmitry Kamenetsky Oct 08 '21 at 13:21
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    I've always loved this question. There are in fact many solutions, but I'll let others have a go at it before I post an answer. Do you allow answers where pieces have one point on the boundary? – Jaap Scherphuis Oct 08 '21 at 13:57
  • By identical do you mean congruent or are you disallowing mirror images? – Jaap Scherphuis Oct 08 '21 at 14:59
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    I dare you to put this into your "additional instructions" when ordering a pizza. – Ian MacDonald Oct 08 '21 at 18:08
  • "No, they can't be identical while half have crust and half don't." – TCooper Oct 08 '21 at 21:40
  • I've added a bonus question for identical pieces. – Dmitry Kamenetsky Oct 08 '21 at 22:47
  • sorry for the confusion, I meant that interior pieces are allowed to have one point on the boundary. – Dmitry Kamenetsky Oct 09 '21 at 10:08
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    By any chance, did you read "Things to make and do in the 4th dimension" by Matt Parker? ;) – QBrute Oct 09 '21 at 12:14
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    Here is the same question from math stack exchange that may be of interest (although without the specification of 12 pieces): https://math.stackexchange.com/questions/481527/slice-of-pizza-with-no-crust – Carmeister Oct 10 '21 at 02:40
  • @QBrute I literally asked the exact same question in a comment before seeing it was posted already ;) – Esther Oct 10 '21 at 07:11
  • I can't post an answer (insufficient reputation), but I do have a somewhat oblique answer, based on an ancient cake-cutting riddle. I propose making three straight cuts to create six identically-shaped pizza slices ... and then one additional cut in the horizontal plane to divide those six into 12. The top six will have topping, the bottom six will not. This might be acceptable if (baked dough === crust) even if not at the edge of the original pizza. – noughtnaut Oct 11 '21 at 12:22
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    I will just drop this here https://www.youtube.com/watch?v=9_sMmuKPrnY&ab_channel=TheAxisofAwesome – Marko Stanojevic Oct 20 '21 at 08:19

3 Answers3

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This paper by Joel Haddley and Stephen Worsley answers a slightly different question - finding monohedral disc dissections where not all pieces touch the centre - but the results generally apply to this problem too.

My favourite answer is this one:

enter image description here

Note that this one has interior pieces that don't even have a single point on the boundary.

There is an infinite family of solutions, that interpolates between sybog64's solution and the one above:

enter image description here

The paper also shows some neat variations of loopy walt's solution:

enter image description here

Jaap Scherphuis
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This is a minor upgrade on @sybog64's answer:

enter image description here

One way of thinking about it is to start with this

enter image description here

configuration and then taking groups of 2 slices and rotating each group by 120°.

Dmitry Kamenetsky
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loopy walt
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  • Just to confirm for your first picture: I could construct it from 18 identical arcs, each with a length of 1/6 the circle's circumference? – Joel Rondeau Oct 08 '21 at 19:32
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    @JoelRondeau yes, exactly. – loopy walt Oct 08 '21 at 19:50
  • This is correct and answers the bonus question. Well done! – Dmitry Kamenetsky Oct 08 '21 at 22:54
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    So assuming I have simple kitchen things, how do I slice my pizza thus? The radial cuts through the center are easy enough (butcher's twine to find the center of the pizza and get the radius. Pick a point on the edge to be a pivot for the folded string, and mark an arc with a pizza cutter from the crust to the center of the pizza. Move the string's pivot to where the crust edge was cut, and repeat until 6 arcs are cut). I don't know how to easily find the other pivot points, though. – Dewi Morgan Oct 08 '21 at 23:36
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    @DewiMorgan I'm not fluent with advanced kitchen implements but geometrically the circle centres for the remaing 6 arcs can be obtained by rotating the centre of the pizza 30° around each of the corners of the large hexagon. – loopy walt Oct 09 '21 at 00:06
  • The "bonus question" is added after this answer has been posted ... what kind of bonus is that. – WhatsUp Oct 09 '21 at 02:17
  • Yes sorry, the original question had a mistake. It said identical when I really wanted to have congruent. Then I realised we can have identical too, so added the bonus question. Anyway both answers are great, but I can only accept one. – Dmitry Kamenetsky Oct 09 '21 at 05:40
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    @loopywalt "then taking groups of 2 slices and rotating each group by 120°" I prefer to think of it as mirroring them, as two pieces form a symmetrical shield shape. With this point of view you don't have to do all 6 pairs at the same time. – Jaap Scherphuis Oct 09 '21 at 08:13
  • Interesting images. What tool did you use to create it? – Jikku Jose Oct 09 '21 at 16:23
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    @JikkuJose https://shapely.readthedocs.io/en/stable/index.html – loopy walt Oct 09 '21 at 17:00
  • This looks to be the same as this image from this answer on math.SE. – robjohn Oct 11 '21 at 09:35
  • @DewiMorgan All you need is a 12-pieces-where-6-pieces-don't-have-any-crust pizza cutter. You don't have one? It's a must for every kitchen. – user253751 Oct 11 '21 at 15:07
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    @user253751 Oh, is THAT what those things are for? Yeah, turns out I've got three of them, but they've wedged the drawer shut, so I can't get to 'em. Guess I'm still looking for other solutions, then... – Dewi Morgan Oct 12 '21 at 00:43
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I haven't found a perfect solution, my pieces are symmetrical but not identical

layout of the cut pizza

the pizza is first cut in 6 pieces following an arcs centered on the corner of the last cut piece with radius same as the pizza. the 6 pieces are then cut along their axis of symmetry to get one crusty and one crustless piece

sybog64
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