Trisecting the Angle/Parabola

Theorem
Let $\alpha$ be an angle which is to be trisected.

This can be achieved by means of a parabola.

However, the points on the parabola that are required for this construction cannot be found by using only a straightedge and compass.

Construction
Let $\angle POQ$ be the angle which is to be trisected.


 * TrisectionOfAngleWithParabola.png

Let the parabola $\mathcal P$ be constructed whose equation is $y = 2 x^2$.

Construct the circle $\mathcal C_1$ whose center is at $O$ and whose radius is $1$.

By Equation of Circle, this has the equation:
 * $x^2 + y^2 = 1$

Let $\mathcal C$ intersect $OQ$ at $A$.

Let $AB$ be constructed parallel to the $x$-axis to intersect the $y$-axis at $B$.

Let $AB$ be bisected at $C$.

Let $CE$ be constructed perpendicular to $AB$.

Let $DE$ be tangent to circle $\mathcal C_1$ at the $y$-axis

Hence $E$ is the intersection of $CE$ and $DE$.

Construct the circle $\mathcal C_2$ whose center is at $E$ which passes through $O$.

Let $F$ be the point at which the circle $\mathcal C_2$ intersects the parabola $\mathcal P$.

Let $FG$ be dropped perpendicular to the $x$-axis.

Let $FG$ intersect the circle $\mathcal C_1$ at $H$.

The angle $\angle POH$ is the required trisection of $\angle POQ$.

Proof
By construction, $E$ is the point $\tuple {\dfrac {\cos \angle POQ} 2, 1}$.

From Equation of Circle passing through Origin, $C_2$ has the equation:

Also see

 * Trisection of Angle by Neusis Construction
 * Trisection of Angle by Archimedean Spiral