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Let's say you have invited $(n-1)$ people for dinner. You decide that the main course consists of one pizza for each guest, so you order $n$ pizzas. Unfortunately, the pizza guy on the scooter trips on his way to your house and loses all but one pizza.

As you don't want to disappoint your guests too much, you try to divide the (round) pizza up into $n$ equal pieces, so everyone gets an equal share. You are, however, not in the company of someone with a compass. You do have an unmarked ruler and a pen that can leave a colored, eatable solution on the pizza. You can also fold the pizza as much as you want. You can use the creases on the pizza that are left one the surface of the pizza after one has unfolded it.

We assume that the pizza is perfectly round.

Question: Is there a method by which we can divide the pizza into any $n$ equal parts using the prescribed materials and rules?

Thanks,

Max

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    By the way, the pizza guy on the scooter is all right after he trips...2011-04-24
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    Should the parts be equal in area or also equal in shape?2011-04-24
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    @joriki I had not thought about that yet! I am curious about both subquestions: the one on which the requirement "the pieces should be equal in shape" is put and the one without that requirement.2011-04-24
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    If your ruler can be used as a circle drawing tool, then the solution below in my answer will work.2011-04-24
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    If I remember right $n=11$ cannot be done with a marked ruler and a compas. So my guess is no, but I don't know for sure if marking would allow one to solve quitic equations. Making marks is more that using a marked ruler.2011-04-24
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    @picakhu Well, the ruler is an unmarked straightedge.. Can an unmarked straightedge be used to draw a circle?2011-04-24
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    Why not? Start with marking it (arbitrary units)2011-04-24
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    @Max: you are right however, it really depends on how you define a ruler. Perhaps you want to make that a bit more clear. Because in my head, ruler always goes with compass..2011-04-24
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    Here's how I interpret the question; @Max, please correct me if I'm wrong: We don't have a compass, so we can't draw arbitrary circles. The ruler is marked and stays unmarked. The pen is for marking the pizza, not for marking the ruler. This is a mathematical problem and not an exercise in lateral thinking, so the idea is not to figure out how to use the given instruments to draw arbitrary circles without a compass, but to decide whether $n$ equal shares can be constructed using only an unmarked ruler and folding. (Note that folding allows us to draw certain circle arcs.) Correct?2011-04-24
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    @ picakhu I wasn't trying to make a point by asking if an unmarked straightedge can be used to draw a circle, I was genuinely curious! I think it's clever of you to find a way to use the straightedge as a compass, but from now one let's asume we can't do it.2011-04-24
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    @ joriki Correct :)!2011-04-24
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    Of course I meant "the ruler is *unmarked* and stays unmarked".2011-04-24
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    @ joriki yes that's how I interpreted your comment.2011-04-24
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    @User9176, that is not correct.2011-04-24
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    This is why you should always invite Godel, he won't eat anything anyway. :-)2011-04-24
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    @Max: OK, then I think the version with equal shapes is impossible for arbitrary $n$. As far as I can see, all we can do through folding is making straight creases between known points, bisecting the line segments between known points, and drawing arcs with the radius of the pizza; and we could have done all those things if we'd had a compass, so I don't think we can do more than we could with an unmarked ruler and a compass. Also I think the parts having the same shape forces them to each include the $n$-th part of the circumference, so we could construct a regular $n$-gon from the solution.2011-04-24
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    There's always $k$ people that are crust hogs, $j$ people that are crust tolerant and $n-k-j$ people that don't eat the crust. Very important!2011-04-25
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    The origami people can solve arbitrary equations of second or third degree by folding alone (according to their rules of folding). So you would have to check for which $n$ the construction of the angle $2\pi/n$ can be reduced to the solution of such an equation.2011-04-25
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    Is it OK if the pieces are (highly) disconnected? eg, a "piece" consists of a bunch of disconnected circles of radii 1/sqrt(n*2^k)2011-05-01
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    Nope sorry Nick these are picky guests...2011-05-02
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    @all this is possibly useful: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.123.6667&rep=rep1&type=pdf .2011-05-19

1 Answers 1

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Quick trip to google brought this up.

Here is evidence that it can be done:

It is possible to do all compass and straightedge constructions without the straightedge. That is, it is possible, using only a compass, to find the intersection of two lines given two points on each, and to find the tangent points to circles. It is not, however, possible to do all constructions using only a straightedge. It is possible to do them with straightedge alone given one circle and its center.

The Circle is our pizza.

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    It is not the answer to the question, but thanks for making me lookt at this beautiful solution.2011-04-24
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    @user9325, the solution is edited. It MAY be a proper solution now.2011-04-24
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    @ picakhu thank you for your answer. It's getting pretty late in the Netherlands at the moment so I'm going to sleep now and read through everything carefully tomorrow.2011-04-24
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    @picakhu: a) That's a passage in a Wikipedia article, not a proof. b) It only means that you can do all constructions of points that way; it doesn't mean you can construct circles with a radius different from the given circle. So you can't use this to construct the nice semicircle solution that you linked to.2011-04-24
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    @joriki, you could give the specifics of the statement on wiki. I believe I took the "it is possible to do ALL ... constructions" a bit too literally.2011-04-24
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    @ picakhu, very nice, but I see that there is a definition problem here. You can construct any *point* you can construct with straightedge and compass, but you cannot draw the circle arches, but you can construct an arbitrary number of points on the circles.2011-04-24
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    However, even with ruler and compass you can't construct e.g. an angle of $2 \pi/7$, so if $n=7$ I don't see how you're going to divide the pizza.2011-04-24
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    @Robert, you do not have to for this problem.2011-04-24
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    Oops, sorry, I should have checked that link2011-04-24
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    @picakhu I'm sorry but I've read all the links and I agree with joriki and user9325, as you need to construct the semicircles in the solution you need to, something that (apparently) can't be done by means of straightedge alone (I can understand your link though, I think the wikipedia article is a bit misleading...)2011-04-25