If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.

Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$. But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$

Actually the original problem is to show that the affine variety $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose, whose polar equation is certainly $r=\sin 2\theta$. (Exercise 7(b), section 1.2, Ideals, Varieties and Algorithms, 3rd edition, David Cox etc.)

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    $\begingroup$ I assume it's the standard parametrization that $r = \sqrt{x^2 + y^2}$. In that case, if you end up with $r^2 = \sin^2{2\theta}$, there isn't any concern of $r$ being negative when you take the square root, since by definition $r$ is positive. Thus $r$ is the positive solution to $\sqrt{\sin^2{2\theta}}$, so indeed $r = \sin{\phi}$ for some $\phi$ (where that $\phi = 2\theta$). If this makes sense, I'd be happy to turn it into a full answer. $\endgroup$ – jamaicanworm Jan 31 '12 at 5:12
  • $\begingroup$ If we use one of the conventions (when $r$ is negative, reflect in the origin), the $r=\pm \sin 2\theta$ just traces out the figure twice. $\endgroup$ – André Nicolas Jan 31 '12 at 5:15
  • $\begingroup$ @jamaicanworm, yes, $r$ must be positive, but not so $\sin2\theta$. The correct solution is $r=|\sin2\theta|$. $\endgroup$ – Gerry Myerson Jan 31 '12 at 5:19
  • $\begingroup$ @Kou: If we decide that when $r<0$, the curve is not defined (which is one convention), then $r=\sin 2\theta$ has two leaves, not four. $\endgroup$ – André Nicolas Jan 31 '12 at 5:26
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    $\begingroup$ @Kou: Sorry for delay, I was away. Make the usual substitution. We get $r^6=4r^4\cos^2\theta\sin^2\theta=r^2\sin^2 2\theta$. Cancel the $r^4$. Technically, we can't do that, for we are missing the solution $r=0$. But $r=0$, $\theta=$ anything is one of the infinitely many addresses of the origin. The origin will turn out to be also on $r^2=\sin 2\theta$, so we will pick it up. If $r$ is allowed to be negative, then $r=\pm|\sin 2\theta|$. The curve $r=|\sin 2\theta|$ traces out the rose. The curve $r=-|\sin 2\theta|$ traces out the same rose. The end. $\endgroup$ – André Nicolas Feb 3 '12 at 5:56

Allowing $r$ to take both nonnegative and negative values, the graphs of $r=\sin2\theta$ and $r=-\sin2\theta$ (on $0\le\theta<2\pi$) are identical—they contain the same set of geometric points in the plane—though they are traced out differently. For $r=\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the first quadrant petal, then the fourth quadrant petal, then the third quadrant petal, and finally the second quadrant petal. For $r=-\sin2\theta$, as $\theta$ increases from $0$ to $2\pi$, the graph traces the third quadrant petal, then the second quadrant petal, then the first quadrant petal, and finally the fourth quadrant petal.

edit Here's what I mean:

animation of the two graphs plotting simultaneously

Since the sets of points in the plane described by $r=\sin2\theta$ and $r=-\sin2\theta$ are identical, giving only one of them is sufficient to describe the same set of points as $(x^2+y^2)^3=4x^2y^2$. That's what lets us go from $r^2=\sin^22\theta$ and its equivalent of $r=\pm\sin2\theta$ to only $r=\sin2\theta$.


If $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then $(x^2+y^2)^3=4x^2y^2$ becomes $$ r^2=\sin^2(2\theta) $$ as you say. However, note that the curve $r=\sin(2\theta)$ is symmetric through the origin since $\sin(2(\theta+\pi))=\sin(2\theta)$, so it is the same as $r=-\sin(2\theta)$. In other words, it doesn't matter whether you take $+$ or $-$.

$\hskip{4cm}$ four-leafed clover

  • $\begingroup$ Nice graphic. May I ask how you generated it? $\endgroup$ – bgins Feb 7 '12 at 21:10
  • $\begingroup$ @bgins: It was generated entirely using Mathematica 8. $\endgroup$ – robjohn Feb 7 '12 at 21:47

On the variety $V\left((x^2+y^2)^3-(2xy)^2\right)$, in polar coordinates we have $$ r^6=(x^2+y^2)^3=(2xy)^2=r^4\sin^2{2\theta} $$ or $$ r^2=\sin^2{2\theta} \quad\implies\quad r=\pm\sin2\theta $$ (we can ignore the "trivial" algebraic solution $r=0$ for all $\theta$ since it is geometrically recovered for $\theta\in\pi\mathbb{Z}$). Now $\sin{2\theta}$ has period $\pi$ (furnishing the same magnitude for $r$ at antipodal points), the curve has point symmetry through the origin, allowing us to discard the sign and consider only $r\in[0,1]$. But $$ \sin^2{2\theta}=\frac{1-\cos{4\theta}}{2} $$ has period $\frac{\pi}{2}$, vanishes at the axes or "cardinal" directions, and has its maxima of $1$ when $\cos{4\pi}=-1$, i.e. at $\theta=\frac{\pi}{4}+k\frac{\pi}{2}$. This explains why the polar equation for the solution can take the forms $$ r=\left|\sin{2\theta}\right| =\left(\sin^2{2\theta}\right)^\frac{1}{2} =\left(\frac{1-\cos{4\theta}}{2}\right)^\frac{1}{2} $$ and why it gives us a four-leaved rose.

Qualitatively, the result is the same with any positive power $r=|\sin{2\theta}|^p$, but the petal thickness and area and the attenuation of the maximum radius all vary inversely with $p$.

Plotted with sage (online), here is the solution $r^2=\sin^2{2\theta}$ (with the power $p=1$) in red, with another curve $r^6=\sin^2{2\theta}$ (with $p=\frac{1}{3}$) in blue to illustrate the effect of $p$.

(polar_plot(abs(sin(2*t))^(2/2), (t, 0, 2*pi), color='red')
+polar_plot(abs(sin(2*t))^(2/6), (t, 0, 2*pi), color='blue')).show()

True solution in blue, simpler curve in red.

Finally, as a refresher, a good way to remember or derive the angle-doubling step above is from the two equations $$ \begin{matrix} \cos^2t&+&\sin^2t&=&1\\ \cos^2t&-&\sin^2t&=&\cos2t. \end{matrix} $$ Adding or subtracting (and dividing by two) then yields $$ \cos^2t = \frac{1+\cos2t}{2} \qquad \text{or} \qquad \sin^2t = \frac{1-\cos2t}{2} $$ respectively.


Avoiding the issue entirely: $$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$ $$r^3 = 2(r\sin\theta)(r\cos\theta)$$

$$x = r\cos\theta$$ $$y = r\sin\theta$$

$$r^3 =2xy$$

$$r = (x^2 + y^2)^{\frac 12}$$

$$(x^2+y^2)^{\frac 32} =2xy$$ $$(x^2+y^2)^3=4x^2y^2$$


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