2.- Let $O X Y Z$ be a trirectangular trihedron with vertex $O$ and edges $X, Y, Z$. On the edge $Z$, a fixed point $C$ is taken such that $OC=c$. On $X$ and $Y$, two variable points $P$ and $Q$ are taken respectively, such that the sum $OP + OQ$ is a given constant $k$. For each pair of points $P$ and $Q$, the four points $O, C, P, Q$ lie on a sphere, whose center $W$ is projected onto the plane $OXY$. Determine the geometric locus of this projection. Also, determine the geometric locus of $\mathbf{W}$.
## Solution

![](https://cdn.mathpix.com/cropped/2024_04_26_2f2bf6346a1193d58cb1g-1.jpg?height=474&width=628&top_left_y=1305&top_left_x=1137)

In the figure, the dashed lines show the circles resulting from intersecting the sphere with the coordinate planes. The projections of the center W of the sphere onto these planes coincide with the centers of these circles (denoted F, G, and H in the figure), and since the trihedron is trirectangular, F, G, and H are at the midpoints of the segments PQ, QC, and CP, which are the diameters of their respective circles.

Parametrizing with the distance $\mathrm{OP}=\lambda$, we have trivially in the reference $\mathrm{OXYZ}$ the following coordinates:

$$
\begin{gathered}
\mathrm{P}(\lambda, 0,0) ; \mathrm{Q}(0, \mathrm{k}-\lambda, 0) ; \mathrm{C}(0,0, \mathrm{c}) ; \\
\mathrm{F}\left(\frac{\lambda}{2}, \frac{\mathrm{k}-\lambda}{2}, 0\right) ; \mathrm{G}\left(0, \frac{\mathrm{k}-\lambda}{2}, \frac{\mathrm{c}}{2}\right) ; \mathrm{H}\left(\frac{\lambda}{2}, 0, \frac{\mathrm{c}}{2}\right) ; \mathrm{W}\left(\frac{\lambda}{2}, \frac{\mathrm{k}-\lambda}{2}, \frac{\mathrm{c}}{2}\right)
\end{gathered}
$$

The locus of $\mathrm{F}$ is the line $\mathrm{x}+\mathrm{y}=\frac{\mathrm{k}}{2}$ in the plane XOY. The locus of W is a line parallel to the former, located in the plane $\mathrm{z}=\frac{\mathrm{c}}{2}$, more specifically, it is the intersection of the planes:

$$
\left\{\begin{array}{r}
x+y=\frac{k}{2} \\
z=\frac{c}{2}
\end{array}\right.
$$