In a plane, there are two perpendicular lines: $g \perp l$, and on the line $g$ there are two points: $A$ and $B$. On the line $l$, take any point $P$ and draw perpendiculars from points $A$ and $B$ to the lines connecting $P$ with the given points. What is the geometric locus of the intersection points $M$ of these perpendiculars, as the point $P$ moves along the line $l$?
I. Solution: It does not harm the generality if we choose the $x$-axis of our coordinate system to be the line $g$, and the $y$-axis to be the line $l$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_aa6a1671e0520afe951cg-1.jpg?height=434&width=572&top_left_y=263&top_left_x=755)

Figure 1

The coordinates of point $A$ are: $(a, 0)$, those of point $B$ are: $(b, 0)$, and those of point $P$ are: $(0, z)$. If either $a$ or $b$ is zero, for example, $a=0$ (i.e., $A$ is exactly the intersection point of the two given lines), then the geometric locus degenerates into a single point; this point is $B$.

For the rest, let us assume that neither $a$ nor $b$ is zero, and for now, exclude the case $z=0$ (i.e., $P$ is not at the origin).

The slope of line $AP$ is $-\frac{z}{a}$; the slope of line $BP$ is $-\frac{z}{b}$. Therefore, the equations of the perpendiculars to $AP$ at $A$ and to $BP$ at $B$ are:

$$
y=\frac{a}{z}(x-a) \quad \text{and} \quad y=\frac{b}{z}(x-b)
$$

Solving this system of equations, we obtain the coordinates of point $M$:

$$
x=a+b, \quad y=\frac{ab}{z}
$$

For every value of $z$ (excluding $z=0$!), there is a corresponding coordinate pair. Since $x$ is independent of $z$, the points $M$ lie on a line parallel to the $y$-axis, whose equation is

$$
x=a+b
$$

It remains to determine whether every point on the line described by equation (1) belongs to the geometric locus. Since $y=\frac{ab}{z}$ implies that for every $y$-coordinate (except $y=0$) there is a corresponding $z$ value (i.e., $z=\frac{ab}{y}$), and thus a corresponding point $P$ on the $y$-axis chosen as line $l$. The excluded case $z=0$ can be interpreted as corresponding to the point at infinity on line (1), while the point at infinity on line $l$ corresponds to the point on line (1) where $y=0$. With this extension, we can now say that the desired geometric locus is the entire line $x=a+b$, which is the line symmetric to $l$ with respect to the midpoint of segment $AB$.

Peter Bonyhárd (Budapest VIII., Apáczai Csere g. IV. o. t.)

II. Solution: Let the intersection point of $g$ and $l$ be $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_aa6a1671e0520afe951cg-1.jpg?height=499&width=531&top_left_y=2057&top_left_x=797)

Figure 2

Triangles $MAP$ and $MBP$ are right-angled, so by the Pythagorean theorem (Figure 2):

$$
AM^2 + AP^2 = MP^2 = BM^2 + BP^2
$$

Thus,

$$
\begin{gathered}
AM^2 - BM^2 = BP^2 - AP^2 \\
\text{But} \quad BP^2 = OP^2 + OB^2, \quad AP^2 = OP^2 + OA^2, \text{ and thus} \\
AM^2 - BM^2 = (OP^2 + OB^2) - (OP^2 + OA^2) = OB^2 - OA^2 = \text{constant.}
\end{gathered}
$$

It is known (see the solution to problem K. M. L. 423, 1952, p. 47) that the geometric locus of points in the plane whose squared distances from two given points differ by a constant is a line perpendicular to the line connecting the given points.

Imre Dominyák (Miskolc, Földes F. g. IV. o. t.)

III. Solution: Denote the foot of the perpendicular from point $M$ to line $g$ by $M_1$ (Figure 2), and two pairs of similar right-angled triangles are formed:

$$
\triangle MMA_1 \sim \triangle AOP \quad \text{and} \quad \triangle MMB_1 \sim \triangle BOP
$$

because the angles marked identically in the figure are right angles.

The ratios of corresponding sides are equal:

$$
\begin{aligned}
& \frac{MM_1}{M_1A} = \frac{AO}{OP} \\
& \frac{MM_1}{M_1B} = \frac{BO}{OP}
\end{aligned}
$$

Dividing (1) by (2), we get:

$$
\frac{M_1B}{M_1A} = \frac{AO}{BO} = \text{constant}
$$

Since the position of $M_1$ does not depend on the point $P$, the geometric locus of points $M$ is a line perpendicular to $g$ passing through $M$. According to (3), point $M$ is the reflection of point $O$ over the midpoint of segment $AB$.

Every point on line $M_1M$ belongs to the geometric locus, because for every $M$, we can find a point $P$ on line $l$ that generates it (using the same construction that generates $M$ from $P$).

Zoltán Heinemann (Pécs, Bányaip. techn. II o. t.)

IV. Solution: $ABMP$ (Figure 3, or $AMBP$ in Figure 2) is a cyclic quadrilateral because the segment $PM$ is seen from points $A$ and $B$ at right angles.

![](https://cdn.mathpix.com/cropped/2024_05_02_aa6a1671e0520afe951cg-2.jpg?height=526&width=571&top_left_y=1813&top_left_x=796)

Figure 3

By Thales' theorem, $PM$ is the diameter of the circumscribed circle, and its center $K$ lies on the line $f$ that perpendicularly bisects segment $AB$ ($f \parallel l$). Point $M$ is always the reflection of point $P$ over the point $K$ moving on line $f$, so the geometric locus of points $M$ is the reflection of line $l$ over line $f$.

Mihály Makkai (Budapest V., Eötvös g. II. o. t.)

V. Solution: Construct the orthocenter $H$ of triangle $APB$ (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_aa6a1671e0520afe951cg-3.jpg?height=501&width=487&top_left_y=126&top_left_x=800)

Figure 4

Quadrilateral $AHBM$ is a parallelogram because, by construction, one pair of opposite sides is perpendicular to $AP$, and the other pair is perpendicular to $BP$. The midpoint of segment $AB$ is point $F$. Point $M$ and point $H$ are symmetric with respect to point $F$. As $P$ moves along line $l$, $l$ remains the altitude of triangle $APB$, so the geometric locus of the orthocenters $H$ is line $l$, and thus the geometric locus of points $M$ is the line symmetric to $l$ with respect to the midpoint of segment $AB$.

Gábor Surán (Budapest V., Eötvös g. III. o. t.)