16. (YUG 2) Determine all the triples $(a, b, c)$ of positive real numbers such that the system
$$
\begin{aligned}
a x+b y-c z & =0, \\
a \sqrt{1-x^{2}}+b \sqrt{1-y^{2}}-c \sqrt{1-z^{2}} & =0,
\end{aligned}
$$
is compatible in the set of real numbers, and then find all its real solutions.
16. Clearly $|x| \leq 1$. As $x$ runs over $[-1,1)$, the vector $u=\left(a x, a \sqrt{1-x^{2}}\right)$ runs over all vectors of length $\alpha$ in the plane having a nonnegative vertical component. Putting $v=\left(b y, b \sqrt{1-y^{2}}\right), w=\left(c z, c \sqrt{1-z^{2}}\right)$, the system becomes $u+v=w$, with vectors $u, v, w$ of lengths $a, b, c$ respectively in the upper half-plane. Then $a, b, c$ are sides of a (possibly degenerate) triangle; i.e, $|a-b| \leq c \leq a+b$ is a necessary condition.
Conversely, if $a, b, c$ satisfy this condition, one constructs a triangle $O M N$ with $O M=a, O N=b, M N=c$. If the vectors $\overrightarrow{O M}, \overrightarrow{O N}$ have a positive nonnegative component, then so does their sum. For every such triangle, putting $u=\overrightarrow{O M}, v=\overrightarrow{O N}$, and $w=\overrightarrow{O M}+\overrightarrow{O N}$ gives a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: $\alpha=\angle M O N=\angle(u, v)$ and $\beta=\angle(u, w)$.
Therefore, all solutions of the system are
$$
\begin{array}{lll}
x=\cos t, & y=\cos (t+\alpha), \quad z=y=\cos (t+\beta), & t \in[0, \pi-\alpha] \quad \text { or } \\
x=\cos t, & y=\cos (t-\alpha), \quad z=y=\cos (t-\beta), & t \in[\alpha, \pi] .
\end{array}
$$