A triangle's side $AB$ is fixed. What geometric place does the vertex $C$ describe if the Euler line of the triangle passes through a fixed point? Let's examine the special case when this fixed point is the midpoint of $AB$?
In our right-angled coordinate system, the $X$-axis is the line $AB$, and the origin is the midpoint of $AB$. The coordinates of point $A$ are $(-a, 0)$, those of point $B$ are $(a, 0)$, and those of the variable point $C$ are $(\xi, \eta)$.

The Euler line of the triangle passes through the centroid of the triangle, the circumcenter (and the orthocenter as well). The coordinates of the centroid $S$ of the triangle are

$$
x^{\prime}=\frac{-a+a+\xi}{3}=\frac{\xi}{3} \quad \text { and } \quad y^{\prime}=\frac{0+0+\eta}{3}=\frac{\eta}{3}
$$

The circumcenter $K$ of the circle lies on the $Y$-axis, so its abscissa $p=0$; let its ordinate be $q$. Then

$$
(0-a)^{2}+(q-0)^{2}=(0-\xi)^{2}+(q-\eta)^{2} ; \quad \text { hence } \quad q=\frac{\xi^{2}+\eta^{2}-a^{2}}{2 \eta}
$$

The equation of the line $KS$ is

$$
y-\frac{\eta}{3}=\frac{\frac{\xi^{2}+\eta^{2}-a^{2}}{2 \eta}-\frac{\eta}{3}}{0-\frac{\xi}{3}}\left(x-\frac{\xi}{3}\right)
$$

Rearranging:

$$
y=\frac{3 a^{2}-3 \xi^{2}-\eta^{2}}{2 \xi \eta}\left(x-\frac{\xi}{3}\right)+\frac{\eta}{3} \cdots
$$

This line passes through the fixed point $\left(x_{0}, y_{0}\right)$ if its coordinates satisfy equation 1), i.e., if

$$
y_{0}=\frac{3 a^{2}-3 \xi^{2}-\eta^{2}}{2 \xi \eta}\left(x_{0}-\frac{\xi}{3}\right)+\frac{\eta}{3}
$$

This equation now represents the relationship between the coordinates $\xi, \eta$ of the variable point, and the equation of the line on which point $C$ must lie. This equation can be rewritten as:

$$
\left(\xi-x_{0}\right) \eta^{2}-2 \xi y_{0} \eta+\xi^{3}-3 x_{0} \xi^{2}-a^{2} \xi+3 a^{2} x_{0}=0 \ldots
$$

Thus, we have obtained the equation of a third-degree curve.

If now $x_{0}=0, y_{0}=0$, then the equation of the geometric locus of point $C$ is:

$$
\xi \eta^{2}+\xi^{3}-a^{2} \xi \equiv \xi\left(\xi^{2}+\eta^{2}-a^{2}\right)=0 \ldots
$$

According to this, the geometric locus consists of two parts: one part has the equation $\xi=0$, which is the $Y$-axis. If point $C$ moves along the $Y$-axis, each $\triangle ABC$ is isosceles, and its Euler line is the $Y$-axis, passing through the fixed point $O$. (In such triangles, the centroid and point $K$ lie on the $Y$-axis!)

The other part has the equation: $\xi^{2}+\eta^{2}-a^{2}=0$. This represents a circle with center $O$ and passing through points $A$ and $B$. If point $C$ is on this circle, the Euler line of $\triangle ABC$ is the line $OC$, which rotates around $O$! ($O$ is the circumcenter of $\triangle ABC$, $C$ is the orthocenter; the centroid lies on $OC$.)

The solutions do not consider the $Y$-axis as part of the geometric locus.