Doing Time In Prison

Question

A prisoner was awoken by an earthquake that caused his wall clock to fall to the floor and rolled in the corner of his cell. When the ground stops shaking, he learns that the clock is still working properly. He can hear it ticking but all he can see now are the hands of the clock. The numbers and markings are printed on the glass cover which was broken from the fall. (see above)

A minute later a jail guard passed by, so the prisoner ask him what time is it. But the jail guard notice the dilemma and just told the prisoner that he will return in an hour and shall give him a key to unlock his shackle (that prevents him to reach the clock) if the prisoner figured out by that time when the earthquake stopped. Another minute passed by, the inspecting warden also notice the prisoner's dilemma when he ask the warden what time is it. The warden just told the prisoner that he will return in an hour and shall unlock his cell door if the the prisoner figured out by then what time he has returned. All the time devices inside the prison are precise and synchronized and alarms at 6 p.m. for bedtime and 6 a.m. for breakfast. How can the prisoner be sure to tell the right (hh:mm) time?

Answer 1

The angle between hour and minute hands is 0 at noon or midnight. Every minute the minute hand advances by 1/60 of a revolution, and the hour hand by 1/(12*60) of a revolution, so the angle between them advances by 11/(12*60) of a revolution. (So every 12/11 of an hour the angle is back to zero.) Right now, the angle is a little short of a right angle or 1/4 of a revolution. If we suppose it's 11/12 of a right angle, which is about right, then the time is about 15 minutes after one of those moments when the angle is zero. So: 12:15, 1:20, 2:26, 3:31, 4:37, 5:42, 6:48, 7:53, 8:59, 10:04, 11:10. (Approximately.)

Which, though?

Well, suppose the prisoner waits about 50 minutes, for the next time the hour and minute hands coincide. And let's suppose the mechanism in the clock is precise enough (even after the beating it's just received) that the prisoner can tell to within a couple of seconds when the hour and minute hands are exactly lined up. Those instants are at 12:00:00, 1:05:27, 2:10:54, 3:16:21, 4:21:49, 5:26:16, 6:32:43, 7:38:10, 8:43:38, 9:49:05, 10:54:32. The angle between hour+minute hand and second hand at that point (measuring clockwise from hour+minute) is respectively 0, 132, 264, 36, 174, 306, 78, 210, 348, 120, 252 degrees. So the prisoner should see which of these is closest to the actual angle when the hour and minute hands coincide; then the earthquake time is whichever of the times listed earlier comes about 50 minutes before that.

The prisoner should

exclude candidate times that aren't actually times when he might have been asleep.

Note:

If we make the assumption that the guard isn't actually totally delinquent in his duties, the only possible answer is that the earthquake was at 5:42am, guaranteeing that when he comes past one hour later the prisoner will be out of his cell and not there to claim the promised reward. But I don't think this is the intended solution.

Answer 2

An analog clock mechanism with only two hands has

an 11-fold symmetry

meaning that every angle (counting from the minute hand to the hour hand in the same direction) repeats exactly that many times during a 12 hour period.

The question then becomes: can the second hand (which is the third hand, really) distinguish between those symmetrical cases?

If all the hands move by "ticks", that is, by 60 discrete movements in a circle, then the answer is

yes. You can tell the actual time in one minute, by observing the position the second hand is in, when the minute hand moves by a tick. That position is 12, and you know the time.

Since that's way too easy, we'll assume all the hands move continuously.

In this case,

it's probably not possible to tell the time. There is definitely a difference between 12:00:00 and 1:05:05, but it's way too small to discern by human eye, especially from a clock that has recently taken a severe hit.

Which brings us to the final option: If the second hand moves by ticks, and the others move continuously, it becomes a bit easier to distinguish the similar looking cases.

It'll still be impossible to be sure between adjacent hours, because the precision required (less than 2 degrees between the relative positions of the hands) is too small to be discerned with any certainty, and as noted earlier, the clock did just take a big hit.

Answer 3

He can figure out if the earthquake happened between 6 AM and 6 PM, or the other way around. A time in the form of $h:m$ looks like $5h+m/12:m$ on an analog clock, so the time must look like $5h+m/12+a:m+a$ in these conditions. The difference ($d$) is still $5h-11m/12$, which means its multiplication by 12 will be $60h-11m$, with $m<60$. Assuming $m$ is an integer, $m=12d$ $mod$ $60/11$ and $h=(12d + 11m)/60$. He can work out $a$ from that and rotate the clock $6a$ degrees counterclockwise.