452. Two points, moving along a circle in the same direction, meet every 12 minutes, with the first point completing a circle 10 seconds faster than the second. What part of the circle does each point cover in 1 second?
$\triangle$ Let's accept the circumference length as a unit. Denote the speeds of the first and second points as $v_{1}$ and $v_{2}$ (in fractions of this unit per second). Then, according to the first condition, $v_{1}-v_{2}=\frac{1}{12 \cdot 60}$, and according to the second condition, $\frac{1}{v_{2}}-\frac{1}{v_{1}}=10$. We obtain the system of equations:

$$
\left\{\begin{array}{l}
v_{1}-v_{2}=\frac{1}{720} \\
\frac{1}{v_{2}}-\frac{1}{v_{1}}=10
\end{array}\right.
$$

To solve this system, we first transform the second equation:

$$
v_{1}-v_{2}=10 v_{1} v_{2}, \quad \frac{1}{720}=10 v_{1} v_{2}, \quad v_{1} v_{2}=\frac{1}{7200}
$$

Next, from the first equation of the system, express $v_{1}$ in terms of $v_{2}$, substitute this expression into the last equation, and solve the resulting quadratic equation. Complete the solution independently.

Answer: $1 / 80$ and $1 / 90$.