In $\triangle ABC$, the base $BC$ is fixed, and vertex $A$ moves on a line parallel to the base $BC$. What is the geometric locus of the center of the Feuerbach circle associated with the triangle?
We choose the $X$-axis of our Cartesian coordinate system to be the line $BC$, with the origin $O$ at the midpoint of $BC$. The coordinates of point $B$ are $(b, 0)$, those of point $C$ are $(-b, 0)$, and those of point $A$ are $(\lambda, h)$, where $\lambda$ is a variable and $h$ is a constant.

![](https://cdn.mathpix.com/cropped/2024_05_02_d74f5055c24f5d4e2e99g-1.jpg?height=436&width=466&top_left_y=267&top_left_x=819)

The Feuerbach circle of the triangle passes through the midpoints of the sides, so it passes through the points

$$
O(0,0), \quad P\left(\frac{-b+\lambda}{2}, \frac{h}{2}\right), \quad Q\left(\frac{b+\lambda}{2}, \frac{h}{2}\right)
$$

Since the Feuerbach circle passes through the origin, its equation can be written in the form

$$
x^{2}+y^{2}+k x-l y=0
$$

The coordinates of points $P$ and $Q$ satisfy this equation:

$$
\begin{gathered}
\left(-\frac{b+\lambda}{2}\right)^{2}+\left(\frac{h}{2}\right)^{2}+k\left(\frac{-b+\lambda}{2}\right)+l \frac{h}{2}=0 \\
\left(\frac{b+\lambda}{2}\right)^{2}+\left(\frac{h}{2}\right)^{2}+k\left(\frac{b+\lambda}{2}\right)+l \frac{h}{2}=0
\end{gathered}
$$

Subtracting the corresponding terms from these two equations, we get:

$$
b \lambda+b k=0, \quad \text { i.e., } \quad k=-\lambda
$$

Substituting the value of $k$:

$$
l=\frac{b^{2}+h^{2}-\lambda^{2}}{2 h}
$$

The coordinates of the center of the Feuerbach circle are:

$$
\xi=-\frac{k}{2}=+\frac{\lambda}{2}, \quad \eta=-\frac{l}{2}=+\frac{b^{2}+h^{2}-\lambda^{2}}{4 h}
$$

Eliminating the parameter $\lambda$, the relationship between the coordinates of the center of the Feuerbach circle is:

$$
\eta=\frac{b^{2}+h^{2}-\frac{\xi^{2}}{4}}{4 h} \quad \text { or } \quad y=\frac{4\left(b^{2}+h^{2}\right)-x^{2}}{16 h}
$$

This results in the equation of a parabola. Therefore, if $A$ traces a line parallel to $BC$, the center of the Feuerbach circle of $\triangle ABC$ traces a parabola whose axis is the perpendicular bisector of $BC$ and whose vertex is above the line if $h>0$. This vertex is located at a distance of $\frac{b^{2}+h^{2}}{4 h}$ from $BC$ on this line. (This point will be the center of the Feuerbach circle when $\triangle ABC$ is isosceles.)

Tibor Berger (State Fáy András Grammar School, 8th grade, Budapest IX.)[^0]

[^0]:    ${ }^{1} \mathrm{~A}$ The Feuerbach circle passes through the foot of the altitude from $A$, $A_{1}$. $O A_{1}$ is a chord of the circle, so the center of the circle lies on the perpendicular bisector of this chord, which is the line $x=\frac{\lambda}{2}$.