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?

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 $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.

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 $7.16$.