1. **State the problem:** We have a broken clock where the minute hand completes one full revolution in 72 minutes instead of the usual 60 minutes, and the hour hand moves normally with a 12-hour period (720 minutes per revolution). Starting at 12:00 PM, both hands are aligned. We want to find after how many minutes the hands will meet again. 2. **Define variables and angular velocities:** - Let $t$ be the time in minutes after 12:00 PM. - The minute hand period is 72 minutes, so its angular velocity is $\omega_m = \frac{2\pi}{72} = \frac{\pi}{36}$ radians per minute. - The hour hand period is 720 minutes, so its angular velocity is $\omega_h = \frac{2\pi}{720} = \frac{\pi}{360}$ radians per minute. 3. **Condition for hands meeting:** The hands meet when their angles differ by a multiple of $2\pi$: $$ |\theta_m - \theta_h| = 2\pi k, \quad k = 1, 2, 3, \ldots $$ Since they start aligned at $t=0$, the first time after $t=0$ they meet again corresponds to $k=1$. 4. **Express angles as functions of time:** $$ \theta_m = \omega_m t = \frac{\pi}{36} t $$ $$ \theta_h = \omega_h t = \frac{\pi}{360} t $$ 5. **Set up the equation for meeting:** $$ \left| \frac{\pi}{36} t - \frac{\pi}{360} t \right| = 2\pi $$ Simplify inside the absolute value: $$ \left| \pi t \left( \frac{1}{36} - \frac{1}{360} \right) \right| = 2\pi $$ Calculate the difference: $$ \frac{1}{36} - \frac{1}{360} = \frac{10}{360} - \frac{1}{360} = \frac{9}{360} = \frac{1}{40} $$ So: $$ \left| \pi t \cdot \frac{1}{40} \right| = 2\pi $$ 6. **Solve for $t$:** $$ \pi \frac{t}{40} = 2\pi $$ Divide both sides by $\pi$: $$ \cancel{\pi} \frac{t}{40} = 2 \cancel{\pi} $$ $$ \frac{t}{40} = 2 $$ Multiply both sides by 40: $$ \cancel{40} \frac{t}{\cancel{40}} = 2 \times 40 $$ $$ t = 80 $$ 7. **Interpretation:** The hands will meet again after 80 minutes. **Final answer:** $$ \boxed{80 \text{ minutes}} $$