Can a regular heptagon be constructed using a compass, straightedge, and angle trisector? 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? 
 A: Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:


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
$$(2-x^2)^3+(2-x^2)^2-2(2-x^2)-1=0$$
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 $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.

A: Although there is a good answer with an explicit geometric construction, there is an alternative (algebraic) approach to it.
More concretely, since we want to divide a circumference in seven equal portions, we are interested in constructing an angle of amplitude $2\pi/7$ radians with the given tools. Notice how an angle is determined by its cosine, so we will try to prove that $\cos(2\pi/7)$ is a constructible number.
It is a well-known fact that the degree of $\mathbb{Q}[\cos(2\pi/p)]$ over $\mathbb{Q}$ is $(p-1)/2$ (if you are curious about this, you can check for an answer in Milne's Fields and Galois Theory, Chapter one, Constructions with straight-edge and compass), so in our case,
$$[\mathbb{Q}[\cos(2\pi/7)]:\mathbb{Q}]]=3$$
so the minimal polynomial of $\cos(2\pi/7)$ over $\mathbb{Q}$ is of degree three; lets say that polynomial is
$$x^3+ax^2+bx+c$$
with $a,b,c\in\mathbb{Q}$. But then we can reduce this cubic via the change of variables $x=(y-a)/3$, for which
$$x^3+ax^2+bx+c=\left(\frac{y-a}{3}\right)^3+a\left(\frac{y-a}{3}\right)^2+b\left(\frac{y-a}{3}\right)+c=\\
=\frac{1}{27}(y^3-3ay^2+3a^2y-a^3)+\frac{a}{9}(y^2-2ay+a^2)+\frac{b}{3}(y-a)+c=\\
=\frac{1}{27}(y^3-3ay^2+3a^2y-a^3+3ay^2-6a^2y+3a^3+9by-9ab+27c)=\\
=\frac{1}{27}(y^3+(3a^2-6a^2+9b)y+27c+2a^3-9ab)=\frac{1}{27}(y^3-py-q)$$
Where the last equation has to be understood as a definition/renaming of the coefficients $p,q\in\mathbb{Q}$.
It is now clear that from the roots of this polynomial we can recover the roots of the minimal polynomial given above, since we stablished a bijection between both sets of solutions. If we prove that these roots are constructible with straight-edge, compass, and angle trisector, we will conclude that so does $\cos(2\pi/7)$.
Lets denote by $\alpha$ the possitive root of $4p/3$, so that $\alpha^2=4p/3$. Let $3\theta$ be the angle such that $\cos(3\theta)=4q/a^3$, and use the angle trisector to construct $\cos(\theta)$. We then have that $\alpha\cos(\theta)$ is a root of $y^3-py-q$. Indeed,
$$(\alpha\cos(\theta))^3-p\alpha\cos(\theta)-q=\alpha\frac{4p}{3}\cos^3(\theta)-p\alpha\cos(\theta)-q=\alpha p(\frac{4}{3}\cos^3(\theta)-\cos(\theta))-q=\\
=\frac{\alpha p}{3}\cos(3\theta)-q=\frac{\alpha p}{3}\frac{4q}{\alpha^3}-q=\frac{4pq}{3\alpha^2}-q=\frac{4pq}{3\cdot4p/3}-q=0$$
Where in the third equality we have used the formula of the cosine of the triple of an angle
$$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$
that you should be able to prove knowing the formula of the cosine of the sum of an angle; just express $3\theta=2\theta+\theta$ and operate.
Finally, if $\alpha\cos(\theta)$ was not such that $\cos(2\pi/7)=(\alpha\cos(\theta)-a)/3$, then the other two roots of the polynomial $y^3-py-q$ are real, and one of them, let us denote it by $y_1$, is such that $\cos(2\pi/7)=(y_1-a)/3$. But if we express $d=\alpha\cos(\theta)$, and factor the last polynomial as
$$y^3-py-q=(y-d)(a'y^2+b'y+c')$$
Then we have that
$$(y-d)(a'y^2+b'y+c')=a'y^3+(b'-a'd)y^2+(c'-b'd)y-dc'=y^3-py-q\Leftrightarrow\\
\Leftrightarrow a'=1,\;b'=d,\;c'-b'd=-p\Leftrightarrow a'=1,\;b'=d,\;c'=d^2-p$$
So the coefficients are all constructible, and in this case, since
$$y_1=\frac{-b'+\sqrt{{b'}^2-4c'}}{2}\quad\text{ or }\quad y_1=\frac{-b'-\sqrt{{b'}^2-4c'}}{2}$$
$y_1$ is also constructible. In the end, $\cos(2\pi/7)=(y_1-a)/3$ is constructible, so the heptagon is constructible with straight-edge, compass, and angle trisector.
PS: There is an alternative, yet valid argument in the famous Galois Theory book, from Ian Stewart; check for Theorem $7.16$.
A: We can actually derive an inscribed regular heptagon from the corresponding regular hexagon. The method is based on Conway's The Book of Numbers.
Begin by constructing the hexagon, labeling its vertices A through F in rotational order. Let O be the center of the circle, which is typically identified during the course of this construction. (If not, then the center is rendered by drawing diagonals between two pairs of opposite vertices and noting their intersection.)
If you did not already do so, construct diagonal AD and its perpendicular bisector. The latter will intersect side BC at G and side EF at H.
Bisect OH to generate point I and then draw IC. The latter segment intersects OD at J such that OJ measures one-sixth of the circle radius.
Draw JG (which has slope $3\sqrt3$) and then use the angle trisector to
find the trisecting ray of angle AJG that is closer to ray AJ. This ray intersects the circumscribed circle at K such that arc AK measures one seventh of the circle. The entire heptagon may then be identified through chord copying.
