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.
Let the parabola $\PP$ be constructed whose equation is $y = 2 x^2$.
Construct the circle $\CC_1$ whose center is at $O$ and whose radius is $1$.
By Equation of Circle, this has the equation:
- $x^2 + y^2 = 1$
Let $\CC$ 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 the circle $\CC_1$ at the $y$-axis
Hence $E$ is the intersection of $CE$ and $DE$.
Construct the circle $\CC_2$ whose center is at $E$ which passes through $O$.
Let $F$ be the point at which the circle $\CC_2$ intersects the parabola $\PP$.
Let $FG$ be dropped perpendicular to the $x$-axis.
Let $FG$ intersect the circle $\CC_1$ at $H$.
The angle $\angle POH$ is the required trisection of $\angle POQ$.
Proof
First, notice that because $A$ lies on $\CC_1$:
- $A = \tuple {\cos \angle POQ, \sin \angle POQ}$
This means:
- $B = \tuple {0, \sin \angle POQ}$
Because $C$ is the midpoint of $AB$:
- $C = \tuple {\dfrac {\cos \angle POQ} 2, \sin \angle POQ}$
Because $D$ lies on $\CC_1$:
- $D = \tuple {0, 1}$
and so:
- $E = \tuple {\dfrac {\cos \angle POQ} 2, 1}$
From Equation of Circle, $C_2$ has the equation:
- $\paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {y -1}^2 = \dfrac {\cos^2 \angle POQ} 4 + 1$
Because $F$ lies on both $\CC_2$ and $\PP$, we can solve for the $x$-coordinate of $F$:
\(\ds \paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {2 x^2 - 1}^2\) | \(=\) | \(\ds \dfrac {\cos^2 \angle POQ} 4 + 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 - x \cos \angle POQ + \dfrac {\cos^2 \angle POQ} 4 + 4 x^4 - 4 x^2 + 1\) | \(=\) | \(\ds \dfrac {\cos^2 \angle POQ} 4 + 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 - x \cos \angle POQ + 4 x^4 - 4 x^2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 x^4 - 3 x^2 - x \cos \angle POQ\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \paren {4 x^3 - 3 x - \cos \angle POQ}\) | \(=\) | \(\ds 0\) |
Thus $x = 0$ or $4 x^3 - 3 x - \cos \angle POQ = 0$.
This confirms what is obvious by the construction, namely that $\CC_2$ and $\PP$ intersect at the origin.
We can also see that the solution we are after must lie in the first quadrant, where $x \ne 0$.
So there must be one positive $x$ such that:
- $4 x^3 - 3 x = \cos \angle POQ$
By the Triple Angle Formula for Cosine $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$, it is clear that one solution that works is:
- $x = \map \cos {\dfrac {\angle POQ} 3}$
Because, by construction:
- there is a vertical line through $F$ and $H$
- $H$ lies on the unit circle $\CC_1$
it follows that:
- $H = \tuple {\cos \dfrac {\angle POQ} 3, \sin \dfrac {\angle POQ} 3}$
Therefore:
- $\angle POH = \dfrac {\angle POQ} 3$
$\blacksquare$
Also see
Historical Note
Use of the parabola to trisect an angle appears to have originated with René Descartes.
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $3$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3$