# Is it possible to divide a circle into $7$ equal “pizza slices” (using geometrical methods)?

Or is it possible to divide a circle into n equal "pizza slices" (I don't know how to call these parts, but I think you'll know what I mean), where n hasn't got a common divider with $360$? Or are the $360$ degrees just "arbitrarily" chosen, in such way that it's also possible to make a system with $7$ "degrees" in a circle?

The main question is actually if it's possible with a ruler and a pair of compasses to divide a circle into any number of slices, and if there's a condition for a number (e.g. a common divider with $360$ as I suggested) so it's possible to slice the circle in such a number of pieces.

• What can you do with a pair of compasses that cannot be done with one? – gt6989b Jul 10 '13 at 21:43
• A pair or compasses is a phrase that just means a compass, and it is correct to say it. – Brian Rushton Jul 10 '13 at 21:47
• It's just like a pair of trousers, I suppose. – Hagen von Eitzen Jul 10 '13 at 21:48
• I wonder: Using Paper Folding Geometry it's known that you can trisect an angle, but is it possible to divide it in 7?(or, in general, in any n/n given by a certain formula?) – Bakuriu Jul 11 '13 at 6:41
• You can't divide an arbitrary angle in 7 equal parts with paper folding unless you use a multifold technique, two or more folds at the same time. But you can divide a full disk that way with just single folds. Dividing a full disk into seven equal parts involves solving a cubic rather than a 7th degree equation, and single folds can do that. – Oscar Lanzi Mar 6 '16 at 15:06

## 4 Answers

The following result answers your question. We can divide a circle into $N$ parts by straightedge and compass if and only if $$N=2^kp_1p_2\dots p_t,$$ where the $p_i$ are distinct Fermat primes. (We can have $k=0$, or $t=0$.)

A Fermat prime is a prime of the form $2^{2^j}+1$. There are only $5$ known Fermat primes: $3$, $5$, $17$, $257$, and $65537$.

Since $7$ is not a Fermat prime, we cannot by straightedge and compass do the division you seek.

Remark: The result was first published by Wantzel. Some people give Gauss credit for the result. Gauss certainly was the first to prove that the circle can be divided into $17$ equal parts by straightedge and compass. He almost certainly knew that any $360^\circ/N$ angle, where $N$ is of the shape described above, is constructible. There is no evidence that he knew that nothing else is.

Put $N=9$. Then $N$ is not of the shape described above, since $3$ occurs twice in the factorization. This shows that the $20^\circ$ angle is not constructible. Since $60^\circ$ is certainly constructible, that shows we cannot trisect the general angle by straightedge and compass.

Many books have proofs of the Wantzel result, for example Allan Clark's Elements of Abstract Algebra.

• in the definition of N, can k be equal to zero? – BNJMNDDNN Jul 10 '13 at 22:00
• He did assert the equivalence holds, but provided no proof of the necessity; according to Morris Kline in his history. – David Mitra Jul 10 '13 at 22:01
• Yes, definitely. And $t$ could be $0$, i.e. $N$ a plain power of $2$. – André Nicolas Jul 10 '13 at 22:01
• Can you add a reference link with a proof that N has to be the shape you described? Thanks already – BNJMNDDNN Jul 10 '13 at 22:03
• Of course, if other methods are allowed, then roots of cubic equations can be constructed and the eptagon becomes constructible. – egreg Jul 10 '13 at 22:08

The degree units have nothing to do with constructability using ruler and compasses.

The regular $n$- gon can be constructed if and only if $n$ is the product of a power of two and zero or more distinct odd primes of the form $p=2^k+1$. There are not so many such primes (called Fermat primes), namely $3, 5, 17, 257, 65537$ - and as far as is known today no more. Thus it is possible to construct the $6$-gon and the $8$-gon, but not the $7$-gon and not the $9$-gon. So I suggest you invite another (single!) friend to your pizza.

Exercise: If $2^k+1$ is prime then $k$ is a power of two.

• So e.g. a 360-gon is possible to construct, and it was known since the arising of the degrees system? And when you say "not so many", do you mean finite? And to say that it's not constructible, is that the same as stating that it's impossible to have a pizza divided into 7 equal pieces (and it can only be infinitely approximated)? – BNJMNDDNN Jul 10 '13 at 21:55
• There are geometric tools other than straightedge and compass. There are certain linkages (generalized compasses) that will do the job. One can also do it by a non-allowed use of the straightedge ("verging*). That goes back to Archimedes, probably, if not earlier. – André Nicolas Jul 10 '13 at 22:12
• @BNJMNDDNN it is not known whether there exist finitely or infinitely many Fermat primes; we know only five Fermat primes, and that's why Hagen von Eitzen said "not so many". See Wikipedia. – Ian Mateus Jul 10 '13 at 22:14
• Sure, with geometrical instruments other than plain straightedge and compass. – André Nicolas Jul 10 '13 at 22:25
• You may also be interested in constructions with compass and markable ruler. The technical term is neusis, also known as verging. If we allow that, the general angle is trisectable, and the regular heptagon (your problem) is constructible. – André Nicolas Jul 10 '13 at 22:32

You're looking for constructible angles; see this link:http://en.wikipedia.org/wiki/Compass-and-straightedge_construction

The only numbers of pizza slices you can create are powers of two times distinct Fermat primes (so 8 times 3 times 17 would work).

Divide it in half, then in half again, and one more time. You now have eight slices.

One slice is the geometer's fee; the remaining seven slices belong to the customer.