The Maclaurin Trisectrix

Justin Seago December 8, 2008 Colin Maclaurin (1698-1746)

• Scottish mathematician

• Entered University of Glasgow at the age of 11 and received his M.A. at 14. [2]

• Elected professor of mathematics at the University of Aberdeen at the age of 19, making him the youngest professor in history until March 2008. [4]

• Made use of the form of a Taylor Series that bears his name in A Treatise on Fluxions (1742) wherein he defended Newton’s calculus. [2] [3] Geometry: Trisecting an Angle

Imagine two lines separated by an arbitrary angle (θ).

Now, our objective is to precisely trisect this angle; θ that is, geometrically divide the angle into three equal angles (θ/3).

If we made a line spanning the angle and divided it a into three equal lengths, would there be three equal angles? θ/3? θ/3? a θ/3?

a Trisection of General Angles

Here GeoGebra is employed to model the attempted trisection.

If θ is taken to be the angle we wish to trisect, and θ = 96.29°, then θ/3 should be approximately 32.01°

As you can see in the screen capture from GeoGebra at right, trisecting a line spanning the arc of θ gives at best a rough approximation of θ/3. Trisection of General Angles

Here is a second attempt.

Although this greatly improves our approximation from the previous method, drawing this accurately is difficult and the resulting angles are still not exactly θ/3. Maclaurin’s Solution

In 1742 Colin Maclaurin discovered a curve, now called the Maclaurin Trisectrix or The Trisectrix of Maclaurin, that provides one way of geometrically trisecting an angle exactly.

Angle trisection was one of the geometric problems of antiquity.

Notice from the figure that GeoGebra computes the angle trisection using the Maclaurin Trisectrix to every available decimal place. It is also much easier to utilize for the trisection than the aforementioned approximations. Geometric Construction

1. Begin with a circle centered at a point C with a radius CO along the x-axis.

2. Bisect the radius CO with a point D and run a vertical line in the y-direction through D.

2. Draw a line OBA from O intersecting the vertical line at B and intersecting the circle at A.

3. On the line segment OA, create a point P that is the same distance from O as the distance AB. That is, place P on OA such that OP = AB. The locus of the point P as A traces the circumference of the circle is the Maclaurin Trisectrix. [1] Deriving Parametric Equations

Let OC = a and AOC = θ, then x = OPcosθ = ABcosθ Also AB = OA – OB

If we let OA = c, then by the law of sines θ 1/a∙sinθ = 1/c∙sin(π – 2θ) 0 c = a[sin(π – 2θ)] ⁄ sinθ = a[sin(π)cos(2θ) – sin(2θ)cos(π)] ⁄ sinθ = a∙sin(2θ) ⁄ sinθ = 2a∙cosθ

Let OB = b, then π - 2θ b∙cosθ = ½a 0 b = ½a∙secθ

Hence, AB = c – b = 2a∙cosθ – ½a∙secθ Deriving Parametric Equations x = ABcosθ = (2a∙cosθ – ½a∙secθ)cosθ = 2a∙cos2θ – ½a = ½a(4cos2θ – 1) y = ABsinθ = (2a∙cosθ – ½a∙secθ)sinθ = 2a∙cos(θ)sin(θ) – ½a∙tanθ = ½a[4cos2(θ)sin(θ) ⁄ sin(θ) – 1]tanθ = ½a(4cos2θ – 1)tanθ

Thus, x = ½a(4cos2θ – 1) y = ½a(4cos2θ – 1)tanθ are the parametric equations of the Maclaurin Trisectrix with respect to θ. Polar & Cartesian Equations

From the parametric equations we may easily deduce a polar equation since x = r∙cosθ 0 r∙cosθ = ½a(4cos2θ – 1) 0 r = ½a(4cosθ – secθ) which is the polar equation of the trisectrix.

To find the Cartesian equation, note that r = √x2 + y2 by the Pythagorean theorem. Also, x = r∙cosθ 5 x ⁄ √x2 + y2 = cosθ. √x2 + y2 = ½a[4x ⁄ √x2 + y2 – √x2 + y2 ⁄ x]

Multiplying both sides by √x2 + y2 we get x2 + y2 = ½a[4x – (x2 + y2) ⁄ x] 0 x3 + xy2 = ½a(4x2 – x2 – y2) 0 x3 + xy2 = ½a(3x2 – y2) which is the implicit Cartesian equation of the Maclaurin trisectrix. The Trisection

Let OC = 1, EOP = θ, and ECP = α, then we must show that θ = ⅓α.

Note that tanα = rsin(θ)/(rcosθ – 1).

Recall from the polar equation of the trisectrix that r = ½a(4cosθ – secθ) which implies that tanα is equal to tan3θ after some difficult algebraic manipulation (I used Maple to verify this).

Thus, θ = ⅓α P

r rsinθ

θ α O C E rcosθ Properties: Anallagmatic

The curve is the same as its inverse: As a Pedal Curve

The Maclaurin Trisectrix is the pedal curve of a parabola y = ax2 with a pedal point mirroring the focus of the parabola Focus across the directrix.

Note the asymptote, which is the same distance (½a) Directrix from the double point of the trisectrix as the focus of the parabola is from its vertex. Pedal Point Area of the Loop 1 x = a 4 cos2 q K1 2 T  1 2    y = a 4 cos q K1 tan q a ,  $FRV U $ FRV U dU 2     x' q = 4 a$cos q sin q T  b a ,FRV U $FRV U dU A = y q x' q dq  a T p    3 a , VLQ U $ VLQ U 1 2   = a 4 cos q K1 tan q $4a$cos q sin q dq  2 p K 3    a , , $    p 3 2 3   = 2 a $ 4 cos q sin q Kcos q sin q tan q dq a is the area of the enclosed region □ p  K 3

p 3 π/3 = 4 a2$ 4 cos2 q sin2 q Ksin2 q dq 0

p ^ 3 O (3/2a,0) 1 1 1 = 4 a2$ 4 1 Ccos 2q 1 K cos 2q K 1 Kcos 2q dq 2 2 2 0

p 3 -π/3 1 1 = 4 a2$ 1 K cos2 2q K C cos 2q dq 2 2 0 References

[1] Basset, A. B. (1901), An Elementary Treatise on Cubic and Quartic Curves, 81, http://books. google.com/books?id=yUxtAAAAMAAJ

[2] J J O'Connor and E F Robertson, Maclaurin biography, http://www-groups.dcs.st-and. ac.uk/~history/Biographies/Maclaurin.html

[3] MacLaurin, Colin (1801), A Treatise on Fluxions: In Two Volumes, http://books.google.com/ books?id=QOg3AAAAMAAJ

[4] McNeill, David (2008), University appoints world's youngest professor, http://www.independent. co.uk/news/world/asia/university-appoints-child-prodigy-worlds-youngest-professor-818776.html