The unit length segment $AB$ moves in the Cartesian coordinate system in such a way that $A$ is always on the $x$-axis, and $B$ is always on the $y$-axis. Write the equation of the curve traced by the feet of the perpendiculars dropped from the origin to $AB$.
It is evident that the curve has only one point on the coordinate axes, the origin $O$. Let $P(x, y)$ be a point on the curve that is not on any coordinate axis. The foot of the perpendicular from $P$ to $OA$ is $T$. Then $OT = |x|$ and $PT = |y|$.

The triangles $OPT$, $BOP$, and $OAP$ are similar because their angles are equal (all three triangles are similar to the triangle $BOA$). Therefore, the ratios of corresponding sides are also equal:

$$
\frac{OP}{PB} = \frac{PT}{OT} = \frac{|y|}{|x|} \quad \text{and} \quad \frac{OP}{AP} = \frac{OT}{PT} = \frac{|x|}{|y|}
$$

From these equalities, we get:

$$
|x| \cdot OP = |y| \cdot PB \text{ and } |y| \cdot OP = |x| \cdot AP
$$

Multiply the first equality by $|x|$, the second by $|y|$, and then add them:

$$
\left(x^{2} + y^{2}\right) OP = |x \cdot y| \cdot (AP + PB)
$$

By the Pythagorean theorem, $OP = \sqrt{x^{2} + y^{2}}$; according to our conditions, $AP + PB = AB = 1$. Substituting these:

$$
\left(x^{2} + y^{2}\right) \sqrt{x^{2} + y^{2}} = |xy|
$$

This is true if and only if

$$
\left(x^{2} + y^{2}\right)^{3} = x^{2} \cdot y^{2}
$$

This equation is also satisfied by the coordinates of the origin, so we have shown that every point on the desired curve lies on the curve given by equation (1).

We will show that if a point $Q$ satisfies the equation, then there is a position of segment $AB$ such that the foot of the perpendicular from the origin to $AB$ is exactly $Q$. If one of $Q$'s coordinates is 0, then by (1) the other is also 0, and the origin clearly serves as a foot of the perpendicular, for example, when $A(1; 0)$ and $B(0; 0)$. If $Q(u; v)$, where $u \cdot v \neq 0$, then let

$$
A\left(\frac{v}{\sqrt{u^{2} + v^{2}}}; 0\right) \quad \text{and} \quad B\left(0; \frac{u}{\sqrt{u^{2} + v^{2}}}\right).
$$

Then the length of $AB$ is 1 by the Pythagorean theorem, and it is easily seen that $OQ$ is perpendicular to $AB$.

This shows that the equation of the desired curve is

$$
\left(x^{2} + y^{2}\right)^{3} = x^{2} + y^{2}
$$

Based on the work of Gyula Pap (Debrecen, Fazekas M. Gymnasium, II. class)
![](https://cdn.mathpix.com/cropped/2024_05_02_658301e3d3a71e15dec3g-1.jpg?height=1034&width=696&top_left_y=1666&top_left_x=693)