Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Euclid has a magical compass with which he can trisect any angle. Together with a regular compass and a straightedge, can he construct a regular heptagon?

share|cite|improve this question
Apparently, you can. (see page 44 onwards). The idea is to start with trisecting the angle of a right triangle with hypotenuse $1$ and one leg of length equal to $\frac1{2\sqrt 7}$. See the link for more details. – J. M. Dec 22 '11 at 15:07

Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:

heptagon via trisection

Draw the circle with center $O$ passing through $A$ and on it find $M$ so that $AM=OA$. Bisect $OM$ at $N$, and trisect at $P$, and find $T$ on $NP$ so that $\angle NAT=\frac13\angle NAP$. $AT$ is the needed side of the heptagon.

To validate Plemelj's construction, we must prove that $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$. Since $2\cos\dfrac{2\pi}{7}=2-\left(2\sin\dfrac{\pi}{7}\right)^2$, it follows from $x=2\cos\dfrac{2\pi}{7}$ being a root of $x^3+x^2-2x-1=0$ that $2\sin\dfrac{\pi}{7}$ is a root of


the other roots being $-2\sin\dfrac{\pi}{7}$, $\pm 2\sin\dfrac{2\pi}{7}$, and $\pm 2\sin\dfrac{3\pi}{7}$. The equation can be factored as

$$\left(x^3+\sqrt 7\left(x^2-1\right)\right)\left(x^3-\sqrt 7\left(x^2-1\right)\right)=0$$

The roots corresponding to the first cubic factor are $2\sin\dfrac{\pi}{7}$, $-2\sin\dfrac{2\pi}{7}$, and $-2\sin\dfrac{3\pi}{7}$. Writing the first cubic factor in the form

$$\left(\frac1{x}\right)^3-\frac1{x}=\frac1{\sqrt 7}$$

and making the substitution $\dfrac1{x}=\dfrac2{\sqrt 3}\cos\,\psi$ yields the equation $\cos\,3\psi=\sqrt{\dfrac{27}{28}}$. The desired root corresponds to the choice

$$\psi=\frac13\arccos\sqrt{\frac{27}{28}}=\frac13\arctan\frac1{3\sqrt 3}$$

which yields

$$2\sin\frac{\pi}{7}\cos\,\psi=\frac{\sqrt 3}{2}$$

From the figure, we have $\angle NAP=\arctan\dfrac1{3\sqrt 3}$, so $\angle NAT=\psi$. We thus have $AT\cos\,\psi=AN=\dfrac{\sqrt 3}{2}OA$, and from this we also have $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.