Astroid
A four-cusped star-shaped hypocycloid
Intermediate (undergraduate level)
Goal of this page
Understand the equation of the astroid and its construction as the path (a hypocycloid) traced when a circle one quarter the size rolls inside another circle.
1. What is an astroid?
The astroid is a four-cusped hypocycloid — a star-shaped curve with $4$ cusps (sharp points where the tangent direction changes abruptly). It is the path traced by a point on a circle of radius $a/4$ as that circle rolls without slipping inside a fixed circle of radius $a$; equivalently, it is the hypocycloid with radius ratio $4:1$.
The name comes from the Greek astron (star), reflecting its star-like shape; it is a different word from asteroid (a minor planet).
2. Equation
Implicit form
$$x^{2/3} + y^{2/3} = a^{2/3}$$Parametric form: $x = a\cos^3 t,\ y = a\sin^3 t,\ t\in[0, 2\pi]$.
Area: $S = \dfrac{3\pi a^2}{8}$ ($\tfrac38$ of the area $\pi a^2$ of the circle of radius $a$). Perimeter: $L = 6a$ (interestingly, unlike the circle, no $\pi$ appears). It has $4$ cusps of length $a$ along the $x$- and $y$-axes.
3. Derivation
3.1 From the rolling circle to the parametric form
Let a circle of radius $r=a/4$ roll without slipping inside a fixed circle (centre at the origin, radius $R=a$) (Fig. 0). The centre $C$ of the rolling circle lies at distance $R-r$ from the origin; writing its angle as $\theta$,
$$C=\bigl((R-r)\cos\theta,\ (R-r)\sin\theta\bigr).$$The contact point currently sits at angle $\theta$ on the large circle, pointing straight outward from $C$ (direction $\theta$). From the condition “rolling without slipping” = the arc lengths in contact are equal on both circles, measured from the initial contact point at $\theta=0$ up to now,
$$\underbrace{R\,\theta}_{\text{arc, large circle}}=\underbrace{r\,\beta}_{\text{arc, small circle}}\ \Longrightarrow\ \beta=\frac{R}{r}\theta.$$$\beta$ is the angle on the small circle from the “contact direction” to the marked tracing point $P$. Rolling on the inside, $P$ lags the contact direction $\theta$ by the angle $\beta$, so the direction of $P$ seen from $C$ is $\gamma=\theta-\beta=-\dfrac{R-r}{r}\theta$. Hence
$$P=C+r(\cos\gamma,\sin\gamma)=\Bigl((R-r)\cos\theta+r\cos\tfrac{R-r}{r}\theta,\ (R-r)\sin\theta-r\sin\tfrac{R-r}{r}\theta\Bigr).$$This is the general hypocycloid. Since the rolling circle has radius $a/4$, we have $R=4r$, so $\dfrac{R-r}{r}=3$, giving
$$x=3r\cos\theta+r\cos3\theta,\qquad y=3r\sin\theta-r\sin3\theta.$$Substituting the triple-angle formulas $\cos3\theta=4\cos^3\theta-3\cos\theta,\ \sin3\theta=3\sin\theta-4\sin^3\theta$, the linear terms in $\cos\theta,\sin\theta$ cancel exactly:
$$x=3r\cos\theta+r(4\cos^3\theta-3\cos\theta)=4r\cos^3\theta,$$ $$y=3r\sin\theta-r(3\sin\theta-4\sin^3\theta)=4r\sin^3\theta.$$Writing $4r=R=a$ and $t=\theta$, we obtain the parametric form (the rotation terms vanish, leaving pure cubes — the essence of the astroid).
Parametric form
$$x=a\cos^3 t,\qquad y=a\sin^3 t,\qquad t\in[0,2\pi].$$3.2 From the parametric form to the implicit form
It suffices to eliminate $t$. From the two equations,
$$\left(\frac{x}{a}\right)^{2/3}=\cos^2 t,\qquad \left(\frac{y}{a}\right)^{2/3}=\sin^2 t.$$Adding these and using the Pythagorean identity $\cos^2 t+\sin^2 t=1$, then multiplying by $a^{2/3}$,
$$\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{a}\right)^{2/3}=1\ \Longrightarrow\ x^{2/3}+y^{2/3}=a^{2/3}.$$3.3 Area $S=\dfrac{3\pi a^2}{8}$
Below we use the derivatives $\dfrac{dx}{dt}=-3a\cos^2 t\sin t,\ \dfrac{dy}{dt}=3a\sin^2 t\cos t$. The enclosed area can be written with Green’s theorem:
$$S=\frac12\oint(x\,dy-y\,dx)=\frac12\int_0^{2\pi}\!\Bigl(x\tfrac{dy}{dt}-y\tfrac{dx}{dt}\Bigr)dt.$$Once again the identity makes the integrand factor neatly:
$$x\tfrac{dy}{dt}-y\tfrac{dx}{dt}=3a^2\cos^4 t\sin^2 t+3a^2\sin^4 t\cos^2 t=3a^2\cos^2 t\sin^2 t\underbrace{(\cos^2 t+\sin^2 t)}_{=1}=3a^2\cos^2 t\sin^2 t.$$By the double-angle formula $\cos^2 t\sin^2 t=\tfrac14\sin^2 2t=\tfrac18(1-\cos4t)$, the integral is
$$\int_0^{2\pi}\cos^2 t\sin^2 t\,dt=\int_0^{2\pi}\tfrac18(1-\cos4t)\,dt=\frac{\pi}{4},$$ $$\therefore\ S=\frac{3a^2}{2}\cdot\frac{\pi}{4}=\frac{3\pi a^2}{8}.$$This is $\dfrac38$ of the area $\pi a^2$ of the inscribed circle of radius $a$. The star fits entirely inside the circle, touching it only at the $4$ cusps (Fig. 2).
3.4 Perimeter $L=6a$
Substituting into the arc-length formula, the same factoring as for the area works:
$$\Bigl(\tfrac{dx}{dt}\Bigr)^2+\Bigl(\tfrac{dy}{dt}\Bigr)^2=9a^2\cos^2 t\sin^2 t(\cos^2 t+\sin^2 t)=9a^2\cos^2 t\sin^2 t,$$ $$\sqrt{\Bigl(\tfrac{dx}{dt}\Bigr)^2+\Bigl(\tfrac{dy}{dt}\Bigr)^2}=3a\,\lvert\cos t\sin t\rvert=\frac{3a}{2}\,\lvert\sin 2t\rvert.$$Because an absolute value appears, we use symmetry: split into $4$ equal arcs at the $4$ cusps, compute one arc where $\sin 2t\ge0$, and multiply by $4$:
$$\ell=\int_0^{\pi/2}\frac{3a}{2}\sin 2t\,dt=\frac{3a}{2}\Bigl[-\tfrac12\cos2t\Bigr]_0^{\pi/2}=\frac{3a}{2},\qquad L=4\ell=6a.$$(If one ignores the absolute value and integrates over $0\to2\pi$, the signs cancel and give $0$. Since arc length is positive, splitting the interval for $\lvert\sin2t\rvert$ is the key.)
4. The sliding ladder and tangents (envelope)
Place a rod (ladder) of length $a$ with its two ends on the $x$- and $y$-axes and slide it. As the rod tilts through every angle, the rods trace out the edge of a star-shaped region that no rod can enter. This boundary — the envelope of the family of rods (the curve tangent to every line of the family) — is the astroid (Fig. 3). Equivalently, it is the boundary swept out by a ladder sliding down a wall.
Why does the star appear? The key is that each sliding rod is a tangent line touching the astroid at exactly one point. Drawing the tangent at the point $(a\cos^3 t,\ a\sin^3 t)$ on the astroid, it cuts the two axes at $(a\cos t,0)$ and $(0,a\sin t)$, and the length between these intercepts is $\sqrt{a^2\cos^2 t+a^2\sin^2 t}=a$ — always equal to the rod length $a$, independent of $t$ (Fig. 4). So “rods of length $a$ with ends on the axes” and “tangents of the astroid” are the same family of lines, and the astroid is the curve they all touch — which is why it appears as the envelope of the rods.
Frequently Asked Questions
Q1. What is an astroid?
A star-shaped curve with 4 cusps. It is the path traced by a point on a circle of radius a/4 as it rolls inside a fixed circle of radius a. Implicitly, x^(2/3)+y^(2/3)=a^(2/3).
Q2. Where does it appear in everyday phenomena?
When a ladder of length a leaning against a wall slides down with its foot sliding out, the envelope of the ladder’s positions is an astroid.
Q3. How do you compute its area?
Integrating with the parametric form (a cos³t, a sin³t) gives S=3πa²/8, which is 3/8 of the area of the circle of radius a.
Q4. Why is it called an astroid?
The name comes from the Greek astron (star), reflecting its star-like shape with 4 cusps. It is a different word from asteroid (a minor planet).
Related topics & references
Related pages on this site
- Cardioid — another roulette curve
- Cycloid — rolling along a straight line
- Circumscribed circle — the fixed circle