Trisectrix of Maclaurin
**TRISECTRIX OF MACLAURIN**

*Balmoral Software*

The lobe of the
Trisectrix of
Maclaurin is a convex x-symmetric droplet-shaped closed curve S with polar
equation
r(t) = 4 cos(t) - sec(t), -π/3 ≤ t < π/3

The curve follows a counterclockwise path as t increases, starting from the cusp
at the origin when t = -π/3, proceeding through Quadrant IV to
the right edge (3,0) at t = 0, then following a symmetric path
through Quadrant I back to the cusp at t = π/3. The maximum
ordinate point of S is

so the
width x height of its bounding rectangle is

### Metrics

The area of
S is

and its perimeter is 8.244653. By
(C2), its centroid abscissa is

### Incircle

The maximum radius of a circle centered on the x-axis and inscribed in S is its
maximum ordinate, so a candidate for the incircle has radius

and center abscissa The
candidate circle must be contained within S, so we require that
c - R and c + R both be within the abscissa range
[0,3] of S, which is true. For verification, we have

### Inellipse

Using z = 3 in Lemma E,
d/dt [x(t) - z]y(t) = d/dt -4sin^{2}(t)tan(t)[2cos(2t) + 1]

has a zero at t* = π/4. The corresponding coordinates are (x*,y*) = (1,1).
We then have

For verification,

### Circumellipse

Using z = 0 in Lemma E,
d/dt [x(t) - z]y(t) = d/dt [4cos(t) - sec(t)]^{2}cos(t)sin(t)

has a zero at

The corresponding coordinates are

We then have

For verification,

### Circumcircle

Since the abscissa extrema are on the x-axis and the bounding rectangle is wider
than it is tall, a candidate for the circumcircle has its radius and center
abscissa both equal to 3/2. For verification, we have

### Summary Table

**Figure** | **Parameters** | Perimeter | Area | Centroid |

Incircle | R = | 7.413905 | 4.374055 | (1.732051,0) |

Inellipse | | 7.826465 | 4.836799 | (1.666667,0) |

Trisectrix | | 8.244653 | 5.196152 | (1.612266,0) |

Circumellipse | | 8.689052 | 5.892208 | (1.535184,0) |

Circumcircle | R = 3/2 | 9.424778 | 7.068584 | (1.5,0) |

The Trisectrix of Maclaurin (red) is a member of a group of similarly-shaped
figures described on these pages, including (inside to outside) the
teardrop curve, the
Tschirnhausen cubic, the
right strophoid and the
piriform curve:

### Relationship with the Folium of Descartes

The Folium of
Descartes coincides with the Trisectrix of Maclaurin after a rotation and
non-uniform scaling of the coordinates. The polar equation of the Folium

is rotated clockwise by π/4, producing the new x-symmetric polar function

Over the lobe domain -π/4 ≤ t < π/4, the maximum abscissa and ordinate
of the rotated Folium are

so to scale the Folium to the comparable coordinate extrema of the trisectrix,
we multiply its abscissas by 3/x_{max} and its ordinates by
producing the coordinate
functions

These coordinate functions cannot be directly compared with the corresponding
ones for the trisectrix since the rates of travel along the path (if t is
considered a time variable) vary between the two curves, and the domains for t
are different. However, the path itself is the same for both curves since the
preceding x(t), y(t) satisfy the
cartesian form
of the trisectrix:
2x(x^{2} + y^{2}) = 2(3x^{2} - y^{2})

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