See main article: Limaçon.
In geometry, a limaçon trisectrix is the name for the quartic plane curve that is a trisectrix that is specified as a limaçon. The shape of the limaçon trisectrix can be specified by other curves particularly as a rose, conchoid or epitrochoid.[1] The curve is one among a number of plane curve trisectrixes that includes the Conchoid of Nicomedes,[2] the Cycloid of Ceva, Quadratrix of Hippias, Trisectrix of Maclaurin, and Tschirnhausen cubic. The limaçon trisectrix a special case of a sectrix of Maclaurin.
The limaçon trisectrix specified as a polar equation is
r=a(1+2\cos\theta)
a
a
-a
\theta=\pi/2
r=a(1+2\cos\theta)
2\pi
\cos\theta
The limaçon trisectrix is composed of two loops.
1+2\cos\theta\ge0
-2\pi/3\le\theta\le2\pi/3
(3a,0)
1+2\cos\theta\le0
2\pi/3\le\theta\le4\pi/3
(a,0)
The curve can be specified in Cartesian coordinates as
a2(x2+y2)=(x2+y2-2ax)2
and parametric equations
x=(a+2a\cos\theta)\cos\theta=a(1+\cos\theta+\cos(2\theta))
y=(a+2a\cos\theta)\sin\theta=a(\sin\theta+\sin(2\theta))
In polar coordinates, the shape of
r=a(1+2\cos\theta)
r=2a\cos(\theta/3)
|a|
a>0
|a|
a<0
r=2a
The inverse of this rose is a trisectrix since the inverse has the same shape as the trisectrix of Maclaurin.
See the article Sectrix of Maclaurin on the limaçon as an instance of the sectrix.
The outer and inner loops of the limaçon trisectrix have angle trisection properties. Theoretically, an angle may be trisected using a method with either property, though practical considerations may limit use.
The construction of the outer loop of
r=1+2\cos\theta
-2\pi/3\le\theta\le2\pi/3
0\le\theta\le2\pi/3
r=2\cos\theta
1
M(1,0)
\theta=\pi/2
A
\overline{AB}
B
(2,0)
\overline{AQ}
\theta=\alpha
\triangle{AQB}
AQ=2\cos\alpha
P
(AQ+1,\alpha)
0<\alpha\le\pi
Given this construction, it is shown that
\angle{QMP}
\angle{PMB}
m\angle{QMB}=2\alpha
\widehat{QB}
r=2\cos\theta
\triangle{AMQ}
\alpha
m\angle{QAB}=m\angle{AQM}=\alpha
\triangle{PQM}
\angle{AQM}
m\angle{PQM}=\pi-\alpha
\angle{QMP}
\angle{QPM}
\alpha/2
m\angle{PMB}=m\angle{QMB}-m\angle{QMP}=2\alpha-\alpha/2=3\alpha/2
\angle{PMB}
m\angle{QMP}/m\angle{PMB}=1/3
m\angle{QPM}/m\angle{QMB}=1/3
m\angle{PAB}/m\angle{QMB}=2/3
The upper half of the outer loop can trisect any central angle of
r=2\cos\theta
0<3\alpha/2<\pi
0<\alpha<2\pi/3
The inner loop of the limaçon trisectrix has the desirable property that the trisection of an angle is internal to the angle being trisected.[5] Here, we examine the inner loop of
r=1+2\cos\theta
\pi\le\theta\le4\pi/3
C
r=1
P
\overline{CM}
M
(1,0)
In Cartesian coordinates the equation of
\overleftrightarrow{CM}
y=k(x-1)
k<0
r= | -k | = |
\sin\theta-k\cos\theta |
-k | |
\cos(\theta-\phi) |
=-k\sec(\theta-\phi)
\tan\phi= | 1 |
-k |
\phi=atan2(1,-k)
Since the normal line to
\overleftrightarrow{CM}
\theta=\phi
\triangle{CAM}
m\angle{CAM}=2\phi
C
(1,2\phi)
With respect to the limaçon, the range of polar angles
\pi\le\theta\le4\pi/3
0\le\theta\le\pi
r=-(1+2\cos(\theta+\pi))=-(1-2\cos\theta)
-\cos(\theta+\pi)=\cos\theta
\alpha
P
-(1-2\cos\alpha)= | -k |
\sin\alpha-k\cos\alpha |
→ (\sin\alpha-k\cos\alpha)-2\cos\alpha\sin\alpha+2k\cos2\alpha=k
→ \cos(\alpha-\phi)-\sin(2\alpha)+2k(
1+\cos(2\alpha) | |
2 |
)=k
→ \cos(\alpha-\phi)-\sin(2\alpha)+k\cos(2\alpha)=0
→ \cos(\alpha-\phi)=\cos(2\alpha-\phi)
The last equation has two solutions, the first being:
\alpha-\phi=2\alpha-\phi
\alpha=0
C
The second solution is based on the identity
\cos(x)=\cos(-x)
\alpha-\phi=\phi-2\alpha
2\phi=3\alpha
m\angle{CAM}=3(m\angle{PAM})
The upper half of the inner loop can trisect any central angle of
r=1
0<3\alpha<\pi
0<\alpha<\pi/3
The limaçon trisectrix
r=a(1+2\cos\theta)
(3a,0)
(a,0)
(3a,0)
3a
Given the limaçon trisectrix
r=1+2\cos\theta
r-1