## 2. Clock

At the moment, according to my watch, it is more than 3:20 but less than 3:25. I observe the exact position of the hands on my watch, and then I move the hands so that the minute hand is in the position previously occupied by the hour hand, and the hour hand is in the position previously occupied by the minute hand.

Can you accurately indicate what time the watch shows at the moment.
2. If $p$ and $g$ are the angles (in radians) that the hour and minute hands form with the direction from the center of the clock face to the 12 o'clock mark, then

$$
g=12(p-h \cdot 2 \pi / 12)
$$

where $h=0,1,2, \ldots$ or 12 is the number of hours ${ }^{1}$. If the angles $p_{0}$

${ }^{1}$ The angles $p$ and $g$ are measured here in the direction of the clockwise rotation; with $0 \leqslant p, g \leqslant 2 \pi, -\text{Note, } \text{pre}$,
and $g_{0}$ characterize the initial positions of the hands, and the angles $p_{1}$ and $g_{1}$ their positions after we have set the clock, then

1) $g_{0}=12\left(p_{0}-\pi / 2\right)$
2) $g_{1}=12\left(p_{1}-2 \pi / 3\right)$
3) $p_{1}=g_{0}, g_{1}=p_{0}$.

## From this

$$
p_{0}=12\left(g_{0}-2 \pi / 3\right)
$$

Substituting this expression into (1), we get

$$
g_{0}=12\left[12\left(g_{0}-2 \pi / 3\right)-\pi / 2\right]
$$

or

$$
143 g_{0}=6 \cdot 17 \pi
$$

But $g_{0}$ is the angle formed with the chosen direction by the minute hand, for which $2 \pi$ corresponds to 60 minutes. Therefore,

$$
g_{0}=\frac{3 \cdot 17 \cdot 60}{143} \text { min. }
$$

Thus, at this moment, the clock shows 3 hours 21 minutes $23(131 / 143)$ seconds $\approx 3$ hours 21 minutes 23 seconds 55 thirds, where 60 thirds $=1$ second.