In the plane, determine the geometric locus of points $A$ for which the altitude of triangle $A B C$ corresponding to $A$ is the geometric mean of the segments $B C + A C$ and $B C - A C$, given that $B$ and $C$ are distinct points.
Solution. Let's set up a Cartesian coordinate system such that the origin coincides with point $B$, the $x$-axis passes through $C$, and we choose the unit so that $C=(1 ; 0)$.

Let the foot of the altitude from $A$ in triangle $ABC$ be $T$. The point $A$ belongs to the desired geometric locus if and only if it is not on the line $BC$ and

$$
A T=\sqrt{(B C+A C)(B C-A C)}
$$

which means

$$
A T^{2}=B C^{2}-A C^{2}
$$

Let's examine when the point $A=(x ; y)$ satisfies these conditions. The coordinates of point $T$ are $(x ; 0)$, so using the known formula for the distance between two points, our condition can be written as

$$
y^{2}=1-\left((x-1)^{2}+y^{2}\right)
$$

Rearranging this, we get

$$
\frac{(x-1)^{2}}{1^{2}}+\frac{y^{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=1
$$

This is the equation of an ellipse with a major axis of $2a=2 \cdot 1=2$, a minor axis of $2b=2 \cdot \frac{1}{\sqrt{2}}=\sqrt{2}$, and its center at point $(1 ; 0)=C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_45b86bca3454e49197b4g-1.jpg?height=344&width=523&top_left_y=1064&top_left_x=796)

Therefore, the geometric locus of the points $A$ is the ellipse whose center is $C$, one endpoint of its major axis is $B$, and the length of its minor axis is $\sqrt{2} BC$, excluding the endpoints of the major axis.