3

Ancient Greeks were not able to trisect a general angle with compass and straightedge: now we know that it is impossible, since we would need to solve a cubic equation while only linear and quadratic equations are soluble.

But Greeks also knew that if the ruler is marked then trisection is possible. My question: which is the set of equations which may be solved with compass and marked ruler?

mau
  • 9,774
  • I think they have to have solutions in field obtained from the rationals by a sequence of quadratic and cubic extensions. But I'm sure this is in the literature, and should be findable by Google. – Gerry Myerson Feb 16 '15 at 12:00
  • http://math.stackexchange.com/questions/145957/on-constructions-by-marked-straightedge-and-compass may be helpful to you. – Gerry Myerson Feb 16 '15 at 12:04
  • 1
    Also please look at Arthur Baragar's paper, Constructions using a compass and twice-notched straightedge, Amer. Math. Monthly 109 (2002), no. 2, 151–164, MR1903152 (2003d:51015). – Gerry Myerson Feb 16 '15 at 22:56
  • Have you had a look at these references? – Gerry Myerson Feb 18 '15 at 02:48
  • Hi Gerry, I already noticed the question here on math.SE, but it seemed to me that it was a bit different, explaining what you could do with an instrument dividing a general angle in $2n+1$ parts. I'm going to search for the paper by Baragar – mau Feb 18 '15 at 06:12
  • (btw, that article is available online, https://faculty.unlv.edu/baragar/papers/monthly151-164.pdf ) – mau Feb 18 '15 at 06:14
  • Had a chance yet to look at the Baragar paper? – Gerry Myerson Feb 20 '15 at 10:33
  • @GerryMyerson yes, eventually I managed. It turned out that with a two-notched straighedge and a compass it is possible to construct points belonging to a field K that lies in a tower of fields for which the index between any two fields is 2,3,5 or 6. This means that 3rd- and 4th-degree equations can be solved and that some quintic may be solved too. Practical problems, like "which regular n-gons may be constructed?", are still open: it is known that a 23-gon cannot constructed, but nothing can still be said about a 11-gon. – mau Feb 20 '15 at 14:55
  • Good. Let me encourage you to post that as an answer (and then to accept it). – Gerry Myerson Feb 20 '15 at 22:51
  • I find it a bit cheating to add my own answer, but since nobody else answered I just added it. – mau Feb 23 '15 at 21:05
  • On the contrary, stackexchange encourages people to answer their own questions (and so do I, as it gives people practice writing things up). – Gerry Myerson Feb 23 '15 at 22:41
  • @mau The question of whether the hendecagon could be constructed with marked ruler was solved affirmatively in 2014. See here. – Parcly Taxel Aug 27 '16 at 15:17

1 Answers1

3

Arthur Baragar studied the constructions where, besides the compass, a straightedge with two notches is available. He shows in this article that with these tools it is possible to trisect an angle and to duplicate the cube, that is to construct a segment whose length is $\sqrt[3]{2}$: these constructions were essentially known to Greek mathematics already.

Baragar gave also the most general answer to the question: for a number $\alpha \in \mathbb{C} $ to be constructable with a straightedge with two notches and a compass, it must belong to a field K that lies in a tower of fields

$$\mathbb{Q} = K_0 \subset K_1 \subset K_2 \subset ... \subset K_n = K$$

for which the index $ [K_j : K_{j-1} ]$ at each step is 2, 3, 5 or 6.

All cubic and quartic equations may be solved, together with some quintic - even some not solvable by radicals; the article states that it is an open problem whether any quintics may be solved.

mau
  • 9,774