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


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.


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 t1 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.


Since the rose is rotationally symmetric with period 2π/3, its circumcircle and circumellipse are the same. The squared-distance function of S
r2(t) = cos2(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
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

Incircle (lobe)R = 1.1592730.106945(0.632303,0)
Inellipse (lobe)a = 0.416666
b = 0.175345
Convex hull5.3196091.610171
CircumcircleR = 16.2831853.141593

Top Page


Copyright © 2021 Balmoral Software ( All rights reserved.