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?

  • $\begingroup$ 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. $\endgroup$ Feb 16, 2015 at 12:00
  • $\begingroup$ math.stackexchange.com/questions/145957/… may be helpful to you. $\endgroup$ Feb 16, 2015 at 12:04
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    $\begingroup$ 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). $\endgroup$ Feb 16, 2015 at 22:56
  • $\begingroup$ Have you had a look at these references? $\endgroup$ Feb 18, 2015 at 2:48
  • $\begingroup$ 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 $\endgroup$
    – mau
    Feb 18, 2015 at 6:12

1 Answer 1


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.


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