The equation of a curve in polar coordinates is given: $r=p \sin 5 \vartheta$ (where $p$ is a given length, $r$ is the radius vector to individual points on the curve, and $\vartheta$ is the corresponding direction angle). Construct the curve and express its equation in Cartesian coordinates.
$1^{\circ}$. If the radius vector $r_{1}$ corresponds to a direction angle $\alpha$ and the radius vector $r_{2}$ corresponds to the angle $\alpha+\pi$, then:

$$
r_{1}=p \sin 5 \alpha \text { and } r_{2}=p \sin (5 \alpha+5 \pi)=-\sin 5 \alpha
$$

Since, however, the rays of the angles $\alpha$ and $\alpha+\pi$ fall on the same line but in opposite directions, $r_{1}$ and $\alpha$ represent the same point as $r_{2}$ and $(\alpha+\pi)$. Therefore, we can construct all points of the curve by varying $\vartheta$ from $0$ to $\pi$.

As $\vartheta$ varies from $0^{\circ}$ to $\pi$, it reaches its absolute maximum five times, specifically when: $\sin 5 \vartheta= \pm 1$, or when $\vartheta=18^{\circ}, 54^{\circ}, 90^{\circ}, 126^{\circ}, 162^{\circ}$, at which points $r=p,-p, p,-p, p$. Thus, the curve has five points in common with the circle drawn from the origin with radius $p$. The other points of the curve lie inside the circle. $p=0$, if $\sin 5 \vartheta=0$, or if $\vartheta=0^{\circ}, 36^{\circ}, 72^{\circ}, 108^{\circ}, 144^{\circ}, 180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0a64cfe771b0ee7e1e4fg-1.jpg?height=594&width=593&top_left_y=527&top_left_x=755)

As $\vartheta$ increases from $0^{\circ}$ to $18^{\circ}$, $r$ increases with $\sin 5 \vartheta$, and when $\vartheta=18^{\circ}$, $r$ reaches its maximum. As $\vartheta$ increases from $18^{\circ}$ to $36^{\circ}$, $\sin 5 \vartheta$ and thus $r$ decrease, and when $\vartheta=36^{\circ}$, $r=0$, or it falls on the origin.

Now let's examine whether there is any relationship between the radius vector corresponding to the direction angle between $0^{\circ}$ and $36^{\circ}$ and the radius vector corresponding to the angle that complements it to $36^{\circ}$.

Let the angle be $\alpha$, and its radius vector be $r_{1}$. The angle that complements it to $36^{\circ}$ is $36^{\circ}-\alpha$, and its radius vector is $r_{2}$. Then $r_{1}=p \sin 5 \alpha, r_{2}=p \sin \left(180^{\circ}-5 \alpha\right)=p \sin 5 \alpha$, or: as $\vartheta$ increases from $0^{\circ}$ to $36^{\circ}$, we obtain a curve that is symmetric about the radius vector corresponding to $\frac{36^{\circ}}{2}=18^{\circ}$. To obtain more points of the curve, let $\vartheta=0^{\circ}, 6^{\circ}, 12^{\circ}, 18^{\circ}, 24^{\circ}, 32^{\circ}, 36^{\circ}$, then $r=0, \frac{p}{2}, \frac{p}{2} \sqrt{3}, p, \frac{p}{2} \sqrt{3}, \frac{p}{2}, 0$. These polar coordinates determine points on the curve $O A$.

As $\vartheta$ increases from $36^{\circ}$ to $72^{\circ}$, $\sin 5 \vartheta$ is always negative, and thus $r$ is also negative. Therefore, if $\vartheta=36^{\circ}, 42^{\circ}, 48^{\circ}, 54^{\circ}, 60^{\circ}, 72^{\circ}$, then $r=0,-\frac{p}{2},-\frac{p}{2} \sqrt{3},-p,-\frac{p}{2} \sqrt{3},-\frac{p}{2}, 0$, and the curve is $B O$.

As $\vartheta$ increases from $72^{\circ}$ to $108^{\circ}$, $\sin 5 \vartheta$ is positive, and thus $r$ is also positive. As $\vartheta$ increases from $108^{\circ}$ to $144^{\circ}$, $\sin 5 \vartheta$ is negative, and thus $r$ is also negative. Finally, as $\vartheta$ increases from $144^{\circ}$ to $180^{\circ}$, $\sin 5 \vartheta$ is positive, and thus $r$ is also positive. Similarly to the previous constructions, we obtain the curves $O C, O D$, and $O E$, each of which has the same properties as the curve $O A$.

Thus, the curve given by the equation is the star-shaped figure $A B C D E O$ within the circle drawn from the origin with radius $p$.

However, it should be noted that $A O B, B O C, C O D, D O E$, and $E O A$ are not arcs of a circle.

$2^{\circ}$. To express the equation of the curve in Cartesian coordinates, we need to use the following relationships between polar and Cartesian coordinates:

$$
r=\sqrt{x^{2}+y^{2}} \text { and } \sin \vartheta=\frac{y}{\sqrt{x^{2}+y^{2}}}
$$

Substituting these values into our given equation and using the easily verifiable

$$
\sin 5 \vartheta=5 \sin \vartheta-20 \sin ^{3} \vartheta+16 \sin ^{5} \vartheta
$$

we get:

$$
\sqrt{x^{2}+y^{2}}=5 p \frac{y}{\sqrt{x^{2}+y^{2}}}-20 p \frac{y^{3}}{\left(\sqrt{x^{2}+y^{2}}\right)^{3}}+16 \frac{y^{5}}{\left(\sqrt{x^{2}+y^{2}}\right)^{5}}
$$

which simplifies to:

$$
\left(x^{2}+y^{2}\right)^{3}=p y\left(5 x^{4}-10 x^{2} y^{2}+y^{4}\right)
$$

or

$$
x^{6}-5 p x^{4} y+3 x^{4} y^{2}+10 p x^{2} y^{3}+3 x^{2} y^{4}-p y^{5}+y^{6}=0
$$

(Szmodics Hildegárd, Kaposvár.)