Trisecting the Angle/Parabola

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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 $\PP$ 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 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 since $A$ lies on $\CC_1$, then $A = \tuple {\cos \angle POQ, \sin \angle POQ}$.

This means that $B = \tuple {0, \sin \angle POQ}$.

Since $C$ is the midpoint of $AB$, we have that $C = \tuple {\dfrac {\cos \angle POQ} 2, \sin \angle POQ}$.

Because $D$ lies on $\CC_1$, then $D = \tuple {0, 1}$ which makes $E = \tuple {\dfrac {\cos \angle POQ} 2, 1}$.

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 use substitution to solve for the x-coordinate of $F$:

\(\displaystyle \paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {2 x^2 - 1}^2\) \(=\) \(\displaystyle \dfrac {\cos^2 \angle POQ} 4 + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 - x \cos \angle POQ + \dfrac {\cos^2 \angle POQ} 4 + 4 x^4 - 4 x^2 + 1\) \(=\) \(\displaystyle \dfrac {\cos^2 \angle POQ} 4 + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 - x \cos \angle POQ + 4 x^4 - 4 x^2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 x^4 - 3 x^2 - x \cos \angle POQ\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \paren {4 x^3 - 3 x - \cos \angle POQ}\) \(=\) \(\displaystyle 0\)

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 also, where $x \neq 0$.

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:

$x = \map \cos {\dfrac {\angle POQ} 3}$

Since, by construction, there is a vertical line through $F$ and $H$ and $H$ lies on the unit circle $\CC_1$, then $H = \tuple {\cos \dfrac {\angle POQ} 3, \sin \dfrac {\angle POQ} 3}$

Therefore, $\angle POH = \dfrac {\angle POQ} 3$



Also see


Historical Note

Use of the parabola to trisect an angle appears to have originated with René Descartes.


Sources