Consider two parallel lines: $e$ and $f$, and point $E$ on $e$, point $F$ on $f$. The lines $e^{\prime}, f^{\prime}$ are formed by rotating $e$ around $E$ by $+\alpha$ degrees, and $f$ around $F$ by $-\alpha$ degrees. What is the geometric locus of the intersection point $M$ of $e^{\prime}$ and $f^{\prime}$, if $\alpha$ takes all values between $0^{\circ}$ and $180^{\circ}$?
If the line $E F$ is perpendicular to $e$ and $f$, then the triangle $E F M$ is isosceles because in this case the rays $E M$ and $F M$ are created by equal and opposite rotations from the rays $E F$ and $F E$. The exception is the case $\alpha = \pm 90^{\circ}$, because then $e$ and $f$ coincide, and $M$ is indeterminate. It is obvious that in this case the geometric locus is the perpendicular bisector of the segment $E F$, except for the midpoint.

![](https://cdn.mathpix.com/cropped/2024_05_02_7dc033280db62c34665ag-1.jpg?height=399&width=748&top_left_y=348&top_left_x=667)

If $E F$ is not perpendicular to $e$ and $f$, then let's take the midpoint of the segment as the origin of our right-angled coordinate system, and let the $X$-axis be parallel to $e$ and $f$. Thus, if the coordinates of $E$ are denoted by $(a, b)$, then $F(-a, -b)$. If $\operatorname{tg} \alpha = m$, then $\operatorname{tg}(-\alpha) = -m$, so the equations of $e'$ and $f'$ are:

$$
y - b = m(x - a), \quad y + b = -m(x + a)
$$

Thus, the coordinates of the intersection point $M$ are:

$$
\bar{x} = -\frac{b}{m}, \quad \bar{y} = -m a
$$

($m \neq 0$, because for $\alpha = 0^{\circ}$ and $180^{\circ}$ there is no intersection point; our calculation is also not valid for $\alpha = \pm 90^{\circ}$, but we know that in this case there is no intersection point because $e' \| f'$.) Eliminating $m$:

$$
m = -\frac{b}{x} = -\frac{\bar{y}}{a}, \quad \text{so} \quad \overline{x y} = a b
$$

Given our assumption, $a \neq 0$; moreover, clearly $b \neq 0$, because $e$ and $f$ do not coincide. Therefore, every point $M$ lies on the hyperbola given by the equation $x y = a b$, whose asymptotes are the coordinate axes, and which passes through $E$. We see that the coordinates of $E$ and $F$ also satisfy the equation (it is even simpler that if $e' \equiv E F$, then $M \equiv F$, and if $f' \equiv E F$, then $M \equiv E$).

We will show that for every point $M$ on this hyperbola, $M E$ and $M F$ form angles $\beta$ and $\gamma$ with $e$ and $f$ respectively, which have equal absolute values but opposite rotational directions. To do this, it suffices to show that $\operatorname{tg} \beta + \operatorname{tg} \gamma = 0$. Indeed, with $\bar{y} = a b / \bar{x}$:

$$
\begin{gathered}
\frac{\bar{y} - b}{\bar{x} - a} + \frac{\bar{y} + b}{\bar{x} + a} = \frac{\frac{a b}{\bar{x}} - b}{\bar{x} - a} + \frac{\frac{a b}{\bar{x}} + b}{\bar{x} + a} = \frac{b(a - \bar{x})}{\bar{x}(\bar{x} - a)} + \frac{b(a + \bar{x})}{\bar{x}(\bar{x} + a)} = \\
= -\frac{b}{\bar{x}} + \frac{b}{\bar{x}} = 0
\end{gathered}
$$

Thus, the geometric locus is given by all points on the hyperbola.

Kiss Tünde (Tamási, Béri Balogh Á. Gimn. III. o. t.)

Remarks. 1. The equation of the hyperbola can also be obtained without writing the equations of $e'$ and $f'$, by noting that if $M'$ and $M''$ are the projections of $M$ onto $e$ and $f$, respectively, then the right-angled triangles $M E M'$ and $M F M''$ are similar, and the ratios of their legs are equal. However, this requires distinguishing several cases depending on which half-plane or strip $M$ is in, and ensuring that the absolute values of the legs are correctly written from the coordinates, and that the rotations of $M' E M$ and $M F M''$ are in opposite directions. If this is neglected, we only get the part of the geometric locus in the first quadrant (if, for example, $a > 0$ and $b > 0$).

These issues also arise when using the right-angled triangles formed by the intersections of $e$ and $f$ with the axes and $M$, by drawing the altitudes from the equal legs of the isosceles triangles. Most of these "second solutions" are incomplete.

2. Some additional, partly elementary solutions tried to infer that we are dealing with a hyperbola from the fact that $M$ and $E$, and $M$ and $F$ are equidistant from the intersections of $e'$ and $f'$ with the axes. This property holds for all hyperbolas, but it should be investigated whether there is no other curve for which a similar equality holds for certain lines. In other words, whether this property is a defining property of the hyperbola.