Equation of Deltoid

Theorem
Let $H$ be the deltoid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the inside of a deferent $C_2$ of radius $a = 3 b$.

Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.

Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.

The point $P = \tuple {x, y}$ is described by the parametric equation:
 * $\begin{cases}

x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$ where $\theta$ is the angle between the $x$-axis and the line joining the origin to the center of $C_1$.

Proof
By definition, a deltoid is a hypocycloid with $3$ cusps.


 * Deltoid.png

By Equation of Hypocycloid, the equation of $H$ is given by:
 * $\begin{cases}

x & = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta} \\ y & = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta} \end{cases}$

From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii, this can be generated by a epicycle $C_1$ of radius $\dfrac 1 3$ the radius of the deferent.

Thus $a = 3 b$ and the equation of $H$ is now given by:
 * $\begin{cases}

x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$