Given a $\Sigma(\alpha)$-system and a line intersecting the axes. Rotate the $\Sigma(\alpha)$-system around the intersection point of the axes. Determine the point in each position of the system whose adjoint line is the intersecting line. What is the geometric locus of these points?

(See the article "On a Geometric Correlation" in the January issue of the problem setter.)
I. solution. Let the axes of the $\Sigma(a)$ system be denoted by $a$ and $b$. We choose the axes of our coordinate system such that its origin coincides with the intersection of the axes of the $\Sigma(a)$ system, and the line $e$ intersecting the axes is perpendicular to the x-axis at a distance $d (\neq 0)$ from the origin. In this coordinate system, the equations of the axes of the $\Sigma(a)$ system are $y = mx$ and $y = nx$; the equation of the line intersecting the axes is $x = d$.

The intersection points of the line $e$ with the lines $a$ and $b$ are:

$$
A(d, md), \quad B(d, nd)
$$

The point $E$ adjoint to $e$ is obtained by constructing the perpendiculars to $a$ at $A$ and to $b$ at $B$, and finding their intersection (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_40c9c1305d8eb1af7359g-1.jpg?height=631&width=442&top_left_y=568&top_left_x=820)

# Figure 1

The equations of these perpendiculars are:

$$
\begin{aligned}
& y - md = -\frac{1}{m}(x - d) \\
& y - nd = -\frac{1}{n}(x - d)
\end{aligned}
$$

During the rotation, $\alpha$ does not change. The tangent of this angle, with direction coefficients $m$ and $n$, can be expressed as follows:

$$
\operatorname{tg} \alpha = \frac{m - n}{1 + m \cdot n}
$$

Equations (1), (2), and (3) hold for every point $E$ generated by the rotation, so this system of equations is the parametric system of equations for the desired geometric locus. In principle, we could express the parameter $m$ from (1) and $n$ from (2), and substitute these values into (3) to eliminate both. However, this would result in a quadratic equation for $m$ and $n$. We can more easily achieve our goal by transforming the equations to express $mn$ and $m-n$ in terms of $x$, $y$, and $d$.

If we subtract (2) from (1) and use known transformations, we get:

$$
mn = \frac{d - x}{d}
$$

If we divide (1) by $n$ and (2) by $m$, and then subtract (2) from (1), after some transformations, we get:

$$
m + n = \frac{y}{d}
$$

We can directly use (4) and (5) if we transform (3) as follows after multiplying by the denominator and squaring:

$$
(1 + mn)^2 \operatorname{tg}^2 \alpha = (m - n)^2 = (m + n)^2 - 4mn
$$

Substituting (4) and (5) and multiplying by $d^2$, we get:

$$
(2d - x)^2 \operatorname{tg}^2 \alpha = y^2 + 4dx - 4d^2
$$

This equation represents a hyperbola; it can be easily transformed into the standard form of a hyperbola. From the equation:

$$
\left(x - 2d \frac{\operatorname{tg}^2 \alpha + 1}{\operatorname{tg}^2 \alpha}\right) \operatorname{tg}^2 \alpha - y^2 = 4d^2 \frac{\operatorname{tg}^2 \alpha + 1}{\operatorname{tg}^2 \alpha}
$$

where

$$
\frac{\operatorname{tg}^2 \alpha + 1}{\operatorname{tg}^2 \alpha} = \frac{1}{\sin^2 \alpha}
$$

it is clear that the center of the hyperbola is on the positive x-axis, at a distance of $\frac{2d}{\sin^2 \alpha}$ from the origin; the length of the real axis, which lies on the x-axis, is $a = \frac{2d \cos \alpha}{\sin^2 \alpha}$, and the length of the imaginary axis is $b = \frac{2d}{\sin \alpha} = a \operatorname{tg} \alpha$. From these, the linear eccentricity is equal to the x-coordinate of the center, so one focus of the hyperbola is at the origin.

Béla Szeidl (Bp. VIII., Apáczai Csere g. III. o. t.)

II. solution. Using the notation from the first solution. The perpendiculars to $a$ and $b$ at points $A$ and $B$ are tangents to a parabola whose focus is the point $O$ and whose directrix is the line $e$. Therefore, the points $E$ see the mentioned parabola at angles $\alpha$ and $(180^\circ - \alpha)$.

From the solution of problem 617 (K. M. L. X. volume, 1st issue), we have proven that the geometric locus of points from which a parabola is seen at angles of $45^\circ$ and $135^\circ$ is a hyperbola. Following the same method as the second solution in that problem (January 1955 issue, p. 25), but using a general angle $\alpha$ instead of $45^\circ$, we arrive at the equation of the geometric locus:

$$
\operatorname{tg} \alpha = \frac{2 \sqrt{y^2 - 2px'}}{2x' + p}
$$

assuming the equation of the parabola is $y^2 = 2px'$. In the present case, the equation of the parabola is $y = -4d(x - d)$, so $p = -2d$ and $x' = x - d$. Substituting these values into (7), we obtain equation (6).

Imre Csiszár (Bp. I., Petófi g. III. o. t.)

III. solution. Using the notation from the first solution, based on the construction, $OBAE$ is a cyclic quadrilateral, and $OE$ is the diameter of the circumscribed circle (since $A$ and $B$ are seen from $O$ at right angles); the center of the circumscribed circle is the midpoint of $OE$: $K$. Draw the central angle corresponding to $\alpha$ and the altitude of $\triangle AKB$, which intersects the line $e$ at the midpoint of the chord $AB$, at point $C$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_40c9c1305d8eb1af7359g-2.jpg?height=521&width=548&top_left_y=1328&top_left_x=775)

# Figure 2

The point $K$ is at a distance $r$ from the point $O$, and its distance from the line $e$ (from $\triangle CAK$) is $KC = r \cos \alpha$. The ratio of these two distances is $\frac{1}{\cos \alpha}$, which is constant and greater than 1. Therefore, the geometric locus of the points $K$ is a hyperbola, one of whose foci is $O$. (See the 268th page of the third-year high school textbook.) Since the points $E$ are in perspective with the points $K$ from $O$ at twice the distance, the geometric locus of the points $E$ is also a hyperbola. The perspective center $O$ is also a focus of this hyperbola.

If $\alpha = 90^\circ$, then the distance $KC = r \cos \alpha$ becomes 0. Therefore, the geometric locus of the points $K$ is the line $e$, and in this case, the geometric locus of the points $E$ is a line parallel to $e$ and at twice the distance from $O$. (This can also be seen from the analytical first solution.)

György Vigassy (Bp. I., Petófi g. IV. o. t.)