Christmas Streak 63/88: Clock #2

Let $\alpha$ be the angle between hour hand and minute hand. Let $s_m$ be angle speed for the minute hand, that is $\frac{360^\circ}{60 minutes} = 6^\circ /minute$ . Let $s_h$ be angle speed for the hour hand, that is $\frac{360^\circ}{12 hours} = \frac{360^\circ}{720 minutes} = 0.5^\circ /minute$ . So, the change of $\alpha$ per minute is

$\Delta s = s_h - s_m = (0.5-6)^\circ /minute = (-5.5)^\circ /minute.$

The negative sign shows that $\alpha$ continues to shrink after 5:00 pm . When 5:00 pm , $\alpha_{5:00} = 150^\circ$ . Let $\alpha_{5:00+n_1} = 110^\circ$ . Since $\Delta s = (-5.5)^\circ /minute$ , then $\alpha_{5:00+n_1} = \alpha_{5:00} + (n_1-1)\cdot(-5.5)$ . We get

$n_1 = \frac {150 + 5.5 - 110}{5.5}.$

Let $\alpha_{5:00+n_2} = -110^\circ$ , that is $110^\circ$ between hour hand and minute hand after $5:00+n_1$ . With the same method to get $n_1$ , we get

$n_2 = \frac {150 + 5.5 + 110}{5.5}.$

Thus, the minutes that he need to stare at the clock is

$n_2 - n_1 = \frac {150 + 5.5 + 110 - 150 - 5.5 + 110}{5.5} = \frac {220}{5.5} = \boxed{40}$