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My question originally rose from a HW exercise in an abstract algebra course, in which I was asked to show that it was impossible to construct a regular heptagon or nonagon.

We have proved in class that it is possible to construct a point $(x,y)\in \mathbb{R}^2$ if and only if the degree of the field extension $[\mathbb{Q}(x,y):\mathbb{Q}]$ is a power of 2. However, I can't manage to use this directly to show that we cannot construct certain polygons. I can show, for example, that the point $(\cos(2\pi/7),\sin(2\pi/7)$ cannot be constructed because of the theorem, but I can't explain why this rules out building some other heptagon which doesn't contain this point.

This link, for example, states that "attempting to construct any polygon is equivalent to finding its internal angles". How can we show that? And how can we show that not being able to construct some point $(\cos \theta,\sin \theta)$ implies that we cannot construct the angle $\theta$ at all?

Dean Gurvitz
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    Not "if and only if". There are non-constructible numbers of degree $4$. See https://math.stackexchange.com/questions/490879/is-the-real-root-of-x43-x-3-constructible – lhf Apr 03 '19 at 11:27
  • Just show that if you could construct any regular heptagon, you could use it to construct the point $(\cos 2\pi/7, \sin 2\pi/7)$. Copying an angle is straightforward, right? – mjqxxxx Feb 01 '24 at 15:21

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Perhaps you know that the 60 degree angle cannot be triseceted. Using the interior angle formula for the regular ngon (n-2)180/n for n=3 we have 60. For n=9 we have 7*20=7*(60/3).

Phillip Feldman
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  • Thank you but that doesn't really answer my question since I'm still unsure how to formally relate the impossibility of constructing certain points to the impossibility of constructing/trisecting angles. – Dean Gurvitz Apr 27 '19 at 17:41
  • (x, y) is a constructible point IFF x and y are both constructible numbers. If points A O and B are constructible then Angle AOB is constructible. – Phillip Feldman Apr 27 '19 at 18:10
  • The problem is that the direction is "if" and not "only if". I know how to show that it is impossible to construct certain points, but that doesn't imply that it is impossible to construct the angle. – Dean Gurvitz Apr 27 '19 at 18:27
  • Take sin or cosine of the angle in question. The angle is constructible IFF the trig value is constructible. To see this, draw the unit circle at the origin. Then sin and cosine correspond to coordinates of one line segment of that angle where the other line is the x axis. – Phillip Feldman Apr 27 '19 at 22:24
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Interesting trivia: Of all the mathematical achievements of the mighty Gauss, he personally considered his 1796 proof of the construction of a regular 17-gon as his most beautiful theorem. Gauss found an unexpected link between contructible polygons and prime numbers. Gauss was so pleased with this result that he requested that a regular 17-gon be inscribed on his tombstone. The stonemason declined, stating that the difficult construction would essentially look like a circle. The discovery ultimately led Gauss to choose mathematics instead of philology as a career

Gauss–Wantzel theorem: A regular $n$-gon (that is, a polygon with n sides) can be constructed with compass and straightedge if and only if

$$ n = 2^a p_1 p_2 p_3 \ldots p_i $$

where $a$ is a positive integers and each $p_i$ is a prime number of the form $2^{2^k} + 1$ (Fermat Prime).

You can find the proofs in this link.

Nilotpal Sinha
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