13.382 Point $P$ is located on the diameter of a circle with radius $R$ between the ends of the diameter $A B$. From this point $P$, three unit masses move in the directions of segments $P A, P B$, and $P C$ such that $P C$ is a half-chord perpendicular to the diameter $\boldsymbol{A} \boldsymbol{B}$. At what distance from $\boldsymbol{A}$ is point $P$ located, given that the velocities are constant and in one unit of time, the first mass reaches point $A$, the second mass reaches point $B$, and the third mass reaches point $C$? The total kinetic energy expended ($\left.\boldsymbol{m}^{2} / 2\right)$ is $a^{2}$ units. Within what range can the value of $a^{2}$ be varied to satisfy the conditions of the problem?
Solution. Let $x=PA$, then $PB=2R-x, PC^2=PA \cdot PB=x(2R-x)$ (PC is the height of the right triangle $\triangle ABC$). Since the masses of the moving points are unitary and their speeds coincide with the corresponding distances $PA, PB$, and $PC$ (points $A, B, C$ are reached in one unit of time), the total energy expended is

$$
\begin{aligned}
& \frac{x^2}{2}+\frac{(2R-x)^2}{2}+\frac{x(2R-x)}{2}=a^2, \text{ hence } \\
& x^2+(2R-x)^2+x(2R-x)=2a^2 \Rightarrow x=R \pm \sqrt{2a^2-3R^2} \Rightarrow \\
& \Rightarrow PA=R+\sqrt{2a^2-3R^2}, PB=R-\sqrt{2a^2-3R^2} \text{ (or vice versa). }
\end{aligned}
$$

The problem makes sense if

$$
R-\sqrt{2a^2-3R^2}>0 \Leftrightarrow\left\{\begin{array}{l}
R^2>2a^2-3R^2 \\
2a^2-3R^2 \geq 0
\end{array} \Leftrightarrow \frac{3}{2}R^2 \leq a^2<2R^2\right.
$$

Answer: $R \pm \sqrt{2a^2-3R^2}$, provided that $\frac{3}{2}R^2 \leq a^2<2R^2$.