COLLINEAR POINTS OF A TRISECTRIX

PAUL SCOTT

University of Adelaide

Introduction

I have been unable to find any reference to the result given below, and believe it is new. I would be glad to hear from any reader who knows of any previous history of this result.

I discovered this result whilst playing with an applet which illustrated the three parallel tangent property of the cardioid.

Let the origin O lie at the cusp of a cardioid, and let points P, R, S be points lying on the cardioid such that segments OP, OQ and OR make equal angles of 120°. Then it is known that the tangents at P, Q and R are parallel for all such positions of the points.

I next extended the applet to include the whole family of limaçons of which the cardioid is a member. I wondered if the parallel tangent property extended to other curves of the family. The applet indicated that the answer is negative here, but I did notice something else of interest.

Discarding the tangents, I noticed that for one particular member of the limaçon family known as the trisectrix, the points P, R and S appeared to be always collinear.

          

The theorem

The general limaçon has polar equation r = 2a cos + k.

When k = 2a, the limaçon is a cardioid. When k = a the limaçon is a trisectrix. This curve has some historical interest in that it can be used to trisect a general angle.

Let the trisectrix have polar equation r = 2 cos + 1. We change the above notation to Q, R, S to avoid using to denote the angle for P !

Theorem Let Q, R, S be three points of the trisectrix which make equal angles of 120° at the origin. Then points Q, R, S are collinear.

Proof Let Q have polar coordinates (r, ). Then Q has Cartesian coordinates
(r cos , r sin ) = (2 cos2 + cos , 2 cos sin + sin ).

Similarly R has Cartesian coordinates (2 cos2 + cos , 2 cos sin + sin ),

and S has Cartesian coordinates (2 cos2 + cos , 2 cos sin + sin ),

where = + 2/3, and = – 2/3 say.

To test whether Q, R, S are collinear, we check that the slopes of segments QR, QS are equal.



Proof of the theorem

A proof is given below. This complicated trigonometric manipulation deserves no marks for elegance, but it does convince me that the result is true! I would be glad to hear from anyone who can find a simple proof of this result.

To test whether Q, R S are collinear, we check that the slopes of segments QR, QS are equal.  That is, we check that

(2 cos sin   +  sin ) – (2 cos sin + sin )

      (2 cos 2   +  cos ) – (2 cos 2   + cos ) 

=  (2 cos sin   +  sin ) – (2 cos sin + sin )

                                              (2 cos 2   +  cos ) – (2 cos 2   + cos )

LHS   =   2 cos sin   +  sin – 2 cos ( + 2/3) sin ( + 2 /3)  – sin ( + 2 /3)

                           2 cos 2   +  cos   –  2 cos 2 ( + 2 /3) – cos ( + 2 /3)

=  4 cos sin   +  2 sin +  [cos + 3 sin ] [3 cos – sin ]  – [3cos – sin ]

                           4 cos 2   +  2 cos   –  [cos + 3 sin )]2  +  [cos + 3 sin )]

=  4 cos sin   +  2 sin +  3 cos2 + 2 cos sin – 3 sin 2  – 3cos + sin

                    4 cos 2   +  2 cos –  cos 2 – 23 cos   sin – 3 sin 2   +  cos + 3 sin

=    3 cos2   +  6 cos sin – 3 sin 2  – 3cos +  3 sin

                  3 cos 2 – 23 cos sin – 3 sin 2   +  3 cos + 3 sin

RHS   =   2 cos sin   +  sin – 2 cos ( – 2 /3) sin ( – 2 /3)  – sin ( – 2 /3)

                              2 cos 2   +  cos    –   2 cos 2 ( – 2 /3) – cos ( – 2 /3)

=  4 cos sin   +  2 sin +  [cos – 3 sin ] [–3 cos – sin ]  + [3cos + sin ]

                             4 cos 2   +  2 cos    –   [cos – 3 sin )]2  +  [cos – 3 sin )]

=  4 cos sin   +  2 sin – 3 cos2 + 2 sin cos + 3 sin 2 + 3cos + sin

                4 cos 2   +  2 cos    –   [cos 2 – 23 sin cos + 3 sin 2 ] +  cos – 3 sin

=    – 3 cos2  + 6 cos sin + 3 sin 2  + 3cos +  3 sin

                  3 cos 2 + 23 cos sin – 3 sin 2   +  3 cos   – 3 sin

Now these two expressions appear to be different, but are they in fact equal?  Let’s check.  We require:

[3 cos2   +  6 cos sin – 3 sin 2  – 3cos +  3 sin ] x

[3 cos 2 + 23 cos sin – 3 sin 2   +  3 cos   – 3 sin ]

= [– 3 cos2  + 6 cos sin + 3 sin 2   + 3cos +  3 sin ] x

[3 cos 2 – 23 cos sin – 3 sin 2   +  3 cos + 3 sin ].

This is true if and only if (ignore the underlining for the moment)

 33cos4 + 6 cos3 sin – 33 cos2 sin 2 + 33cos3 – 3 cos2 sin

        + 18 cos3 sin + 123 cos2 sin 2 – 18 cos sin3 + 18 cos2 sin – 6v3 cos sin 2

– 33 cos2 sin 2   – 6 cos q sin3   + 3 3sin4 – 33 cos sin 2 + 3 sin3

– 33cos3 – 6 cos2 sin + 33 cos sin2 – 33 cos2 + 3 cos sin

+ 9 cos2 sin + 63 cos sin 2 – 9sin3 + 9 cos sin – 33 sin2

=  – 33cos4 + 6 cos3 sin + 33 cos2 sin 2 – 33cos3 – 3 cos2 sin

        + 18 cos3 sin – 123 cos2 sin 2 – 18 cos sin3 + 18 cos2 sin + 63 cos sin 2

+ 33 cos2 sin 2   – 6 cos q sin3  – 3 3sin4 + 33 cos sin 2 + 3 sin3

+ 33cos3 – 6 cos2 sin – 33 cos sin2 + 33 cos2 + 3 cos sin

+ 9 cos2 sin – 63 cos sin 2 – 9sin3 + 9 cos sin + 33 sin2

Deleting the underlined terms which are the same on both sides, that is, if and only if

33cos4 – 33 cos2 sin 2 + 33cos3 + 123 cos2 sin 2 – 63 cos sin 2

          – 33 cos2 sin 2 + 3 3sin4 – 33 cos sin 2 – 33cos3 + 33 cos sin2

– 33 cos2 + 63 cos sin 2 – 33 sin2

=   – 33cos4 + 33 cos2 sin 2 – 33cos3 – 123 cos2 sin 2 + 6v3 cos sin 2

+ 33 cos2 sin 2 – 3 3sin4 + 33 cos sin 2 + 33cos3

– 33 cos sin2 + 33 cos2 – 63 cos sin 2 + 33 sin2 .

Now, again ignoring the underlining,   

3cos4 – 3 cos2 sin 2 + 3cos3 + 43 cos2 sin 2 – 23 cos sin 2 – 3 cos2 sin 2

+ 3sin4 – 3 cos sin 2 – 3cos3 + 3 cos sin2 – 3 cos2 + 23 cos sin 2

– 3 sin2   = 0.

Cancelling the underlined terms, that is, 

cos4 –  cos2 sin 2 + 4 cos2 sin 2 – cos2 sin 2 + sin4   – cos2 – sin2   = 0

or,     cos4 + 2 cos2 sin 2 + sin4   – cos2 – sin2   = 0,

or    (cos2 + sin2 ) 2 – 1  = 0, which is true.

Hence segments QR, QS have equal slopes, and points Q, R, S are collinear as required.


An alternative proof

Viewer Ross Brown from Ontario, Canada has given a nice alternative proof of this result using distances rather than slopes. His proof can be seen here.


Reference

For the context of the discovery of this result, see

http://paulscott.info/DC/limacon/limacon-expl.html


6 October, 2008


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