The segments $A B$ and $B C$, connected at a hinge, have their $A$ endpoint fixed, while $B$ and $C$ can take any position in the plane where the three points are not collinear. What is the geometric locus of the centers of the circles determined by the three points?
A $B$ and $C$ points can rotate around $A$, so if a point $P$ is one point of the geometric locus we are looking for, then every point on the circle with radius $AP$ centered at $A$ is also a point of the geometric locus.

![](https://cdn.mathpix.com/cropped/2024_05_02_01c476e3309675dd27f7g-1.jpg?height=699&width=816&top_left_y=266&top_left_x=641)

Let's examine what the necessary condition is for a point $P$ to belong to the geometric locus. Clearly, it must be possible to construct a triangle with the given segments $AB$ and $BC$ within the circle of radius $AP$ centered at $P$. That is, $AB$ and $BC$ must be chords of the circle with radius $AP$, and thus

$$
AB \leq 2AP \quad \text{and} \quad BC \leq 2AP
$$

must hold.

For given segments $AB$ and $BC$, the geometric locus includes every point in the plane whose distance from $A$ is greater than or equal to the larger of the segments $AB/2$ and $BC/2$.

We can construct a triangle for any point $P$ in the plane if $AP \geq \max(AB/2, BC/2)$ as follows: draw a circle with radius $AP$ centered at $P$, which will be the circumcircle of the desired triangle (with $P$ as the center). Measure the segments $AB$ and $BC$ along the circumference of the circle starting from $A$. The triangle will always be formed, except when $AB = BC = 2AP$, in which case $A \equiv C$, and thus no triangle is formed.