Let's determine the geometric locus of the vertices of the parabolas given by the equation

$$
y=x^{2}-\frac{4 m}{1+m^{2}} x+\frac{1+4 m^{2}-m^{4}}{\left(1+m^{2}\right)^{2}}
$$

if \( m \) takes all real values.
For every real $m$, we get a parabola, because the denominators that appear are always positive and the coefficient of $x^{2}$ is not 0. By completing the square for the first two terms on the right-hand side of (1), we transform it into:

$$
y=\left(x-\frac{2 m}{1+m^{2}}\right)^{2}+\frac{1-m^{4}}{\left(1+m^{2}\right)^{2}}
$$

The last term can be simplified by $(1+m^{2})$, so we further get:

$$
y-\frac{1-m^{2}}{1+m^{2}}=\left(x-\frac{2 m}{1+m^{2}}\right)^{2}
$$

This means that equation (1) can be derived from the equation of the standard parabola $y=x^{2}$ by substituting $x-u$ for $x$ and $y-v$ for $y$, where

$$
\begin{aligned}
& u=\frac{2 m}{1+m^{2}} \\
& v=\frac{1-m^{2}}{1+m^{2}}
\end{aligned}
$$

This implies that the vertex of the standard parabola, the origin, has moved to the point with abscissa $u$ and ordinate $v$, i.e., the vertex of our parabola is the point $C(u, v)$.

As $m$ takes on all real values, the geometric locus of $C$ is obtained by eliminating $m$ from the system of equations (2) and (3). From (3), considering (2), we have

$$
v=1-\frac{2 m^{2}}{1+m^{2}}=1-m u
$$

thus

$$
m=\frac{1-v}{u}
$$

provided $u \neq 0$, which, according to (2), holds for all $m \neq 0$. Postponing the case $m=0$, we substitute (4) into the expression for $m^{2}$ derived from (3):

$$
m^{2}=\frac{1-v}{1+v}=\frac{(1-v)^{2}}{u^{2}}, \quad \frac{1}{1+v}=\frac{1-v}{u^{2}}
$$

since what we simplified by: $1-v=\frac{2 m^{2}}{1+m^{2}}$ only becomes 0 when $m=0$. Finally, rearranging, we get the following relationship between the coordinates of $C$:

$$
u^{2}+v^{2}=1, \text { where, however, } u \neq 0
$$

Thus, if $m \neq 0$, then $C$ lies on the unit circle centered at the origin, but not on the y-axis.

In the remaining case $m=0$, from (2) we get $u=0$ and from (3) we get $v=1$, so the vertex of the parabola is at the point on the positive y-axis of the unit circle.

![](https://cdn.mathpix.com/cropped/2024_05_02_3ce8dad81d9f6f255751g-1.jpg?height=588&width=585&top_left_y=2015&top_left_x=759)

We will show that, except for the point $(0, -1)$ on the unit circle, for every remaining point $C(u, v)$—where $u^{2}+v^{2}=1$ but $v \neq -1$—there is a value of $m$ such that the right-hand sides of (2) and (3) give $u$ and $v$, respectively. This value cannot be anything other than what is given by using (3) and (2):

$$
\begin{aligned}
& m^{2}=\frac{1-v}{1+v}, \quad 1+m^{2}=\frac{2}{1+v} \\
& m=\frac{u}{2}\left(1+m^{2}\right)=\frac{u}{1+v}
\end{aligned}
$$

we only need to consider that the calculations are valid for all $v \neq -1$. We will only show the calculation for (2):

$$
\begin{gathered}
\frac{2 m}{1+m^{2}}=\frac{\frac{2 u}{1+v}}{1+\frac{u^{2}}{(1+v)^{2}}}=\frac{2 u(1+v)}{(1+v)^{2}+u^{2}}= \\
=\frac{2 u(1+v)}{1+2 v+\left(u^{2}+v^{2}\right)}=\frac{2 u(1+v)}{2+2 v}=u
\end{gathered}
$$

Thus, the geometric locus of the vertices of the parabolas (1) is the unit circle centered at the origin, excluding the lowest point $(0, -1)$.

Dombovári Tamás (Budapest, I. István Gimn. IV. o. t.)
Gehér Klára (Szeged, Radnóti M. Gimn. IV. o. t.)
Tóth András (Pápa, Türr I. Gimn. IV. o. t.)

Remarks 1. Those who are familiar with the often used identities

$$
\sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}, \quad \cos x=\frac{1+\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}
$$

immediately see that with the notation $\operatorname{tg} \frac{\varphi}{2}=m$, from (2) and (3) we get $u=\sin \varphi, v=\cos \varphi$, where $\varphi$ is the angle of the ray $O C$ measured from the y-axis in the negative rotational direction. As $m$ increases and runs through the real numbers,

$$
-\frac{\pi}{2}<\operatorname{arctg} m=\frac{\varphi}{2}<\frac{\pi}{2}
$$

$C$ runs through the unit circle, except for the point $\varphi=\pi$.

2. The coordinates of the vertex can also be obtained by considering the parabola as the image of the function (1), finding the abscissa of the minimum point by differentiation, and then determining the ordinate from (1). However, such solutions barely dealt with the re-interpretation (i.e., the rephrasing of the geometric question as a function analysis question).