Rose
**ROSE**

*Balmoral Software*

A three-lobed rose with a petal on the positive x-axis is a closed curve S with
polar equation
r(t) = cos(3t), 0 ≤ t < π

This curve is also known as the
trifolium. Movement
along S is always counterclockwise, starting at its maximum abscissa point (1,0)
and crossing the origin three times. There are two minimum-abscissa points
the lower of which occurs at

so S is non-convex by the multiple
local extrema test. The ordinate extrema of S are

so the width x height of its bounding rectangle is

### Symmetry

Since
r(t + 2π/3) = cos(3t + 2π) = r(t),

the condition (R2) of Lemma R is
satisfied and the rose is rotationally symmetric with period 2π/3.
### Metrics

The perimeter of the rose is 6.682447 and its area is π/4.
### Convex Hull

The convex hull is created by connecting the minimum abscissa points with a
vertical line segment of length
and then repeating the pattern by rotating around the origin at 2π/3
intervals. The far end of the lower-left rose petal is reached at
t = π/3, so the rounded edge of the petal that occurs for values
of t between t_{1} and π/3 is part of the convex hull, and there are
6 such parts by symmetry. We have
r'(t) = -3sin(3t),

so by (L2), the perimeter of the convex
hull is

which is about 20% shorter than that of the rose.
The line segments of the convex hull create three isosceles triangles with the
origin, each having an area of

as shown in blue in the left diagram below. By
(A2), the area of the convex hull is

which is a little over twice the area of the rose.
### Circumconics

Since the rose is rotationally symmetric with period 2π/3, its circumcircle
and circumellipse are the same. The squared-distance function of S
r^{2}(t) = cos^{2}(3t)

is maximized at 1, so that is the circumradius.
### Incircle (lobe)

Consider the x-symmetric right lobe of the rose, where -π/6 ≤ t < π/6.
The maximum ordinate point on this lobe is (c,R), where

and

The largest circle centered on the x-axis and inscribed in the lobe has radius R
and center abscissa c. The circle must be contained within the lobe, so we
require that c - R and c + R both be within the
abscissa range [0,1] of the lobe, which is true. For verification, we have

### Inellipse (lobe)

Using z = 1 in Lemma E,
d/dt [x(t) - z]y(t) = d/dt [6sin(2t) - 5sin(4t) + sin(8t)]/8

has a zero at

The corresponding coordinates are

We then have

For verification, we have

### Summary Table

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

Incircle (lobe) | R = | 1.159273 | 0.106945 | (0.632303,0) |

Inellipse (lobe) | a = 0.416666 b = 0.175345 | 1.937955 | 0.229526 | (0.583334,0) |

Rose | | 6.682447 | 0.785398 | |

Convex hull | | 5.319609 | 1.610171 |

Circumcircle | R = 1 | 6.283185 | 3.141593 |

Top Page

Home

Copyright © 2021 Balmoral Software (http://www.balmoralsoftware.com). All
rights reserved.