Adva van $A B C$ háromszögnek egyik oldala $A B$. Emeljünk a $B C$ oldalra, ennek középpontjában merólegest, mely $A C$-t $M$-ben metszi. Mi az $M$ pont mértani helye, ha a háromszög $C$ csúcsa $A$ körül forog?

Given that one side of triangle $ABC$ is $AB$. Let's raise a perpendicular from the midpoint of side $BC$, which intersects $AC$ at point $M$. What is the geometric locus of point $M$ if the vertex $C$ of the triangle rotates around $A$?
I. Solution. Let $A B C$, then

$$
M C-M A=A C=\text { const. }
$$

thus the geometric locus is a hyperbola.

If $A B=A C$, then the point $M$ coincides with $A$.

(Krisztián György, Pécs.)

II. Solution. Let $A B=c, A C=b$; the origin of the coordinate system is the midpoint $D$ of the line $A B$; the coordinates of point $M$ are $D E=x, M E=y$.

From the right triangle $A E M$:

$$
\overline{A M^{2}}=y^{2}+\left(\frac{c}{2}-x\right)^{2}
$$

From the triangle $M E B$:

$$
\overline{M B^{2}}=y^{2}+\left(\frac{c}{2}+x\right)^{2}
$$

Furthermore,

$$
M B=M C=b-A M
$$

Substituting (1) into (3), and then this result into (2):

$$
\left[b-\sqrt{y^{2}+\left(\frac{c}{2}-x\right)^{2}}\right]^{2}=y^{2}+\left(\frac{c}{2}+x\right)^{2}
$$

After performing the indicated operations and rearranging the equation, we obtain:

$$
4\left(b^{2}-c^{2}\right) x^{2}+4 b^{2} y^{2}=b^{2}\left(b^{2}-c^{2}\right)
$$

or

$$
\frac{x^{2}}{b^{2}}+\frac{y^{2}}{b^{2}-c^{2}}=\frac{1}{4}
$$

The sought geometric locus is an ellipse if $b>c$; a hyperbola if $b<c$.

(Filkorn Jenő, Nyitra.)

The problem was also solved by: Czank K., Freibauer E., Kerekes T., König D., Krausz B., Lukhaub Gy., Lupsa Gy., Rozlosnik P., Sasvári G., Weisz J.