I know that the hour and minute hands of the clock overlap at 12 and once every housr after that, but does the second hand ever overlap the other hands exactly (other than at 12)?

Ahh, this is quite an interesting problem It turns out that the hour and minute hands don''t overlap every hour. The reason is because as the minute hand rotates, so does the hour hand. So if both hands start at 12, by the time the minute hand reaches 12 again, an hour would have passed, but the actually hour hand would be on 1, so they wouldn''t have overlapped yet!

humm...shall we write an equation?

There is a way to do this problem the "right" way, but then there is a way that I am about to argue. Assume it is 12 now. In one hour, the hour hand moves 1/12 of the way through while the minute hand has moved one revolution. When the minute hand is now at the 1 o''clock position, the hour hand has moved 1/144 of the way through, and so on. This is an infinite geometric sum, and you get that the sum is equal to 1/11. So really, you are going 12/11th of an hour. My other way of thinking about this problem was that this event happens 11 times in 12 hours, so you divide out and get the same answer. As for the three handed problem, we can look at it in a similar fashion: The second hand and the minute hand will be in the same position every 60/59 minutes, and the minute and hour hand every 12/11 hours ( 720 / 11 minutes). Hmmmm.... gotta cut this one short but I will get back with the rest of my explanation (I think I am leading toward it not happening because the period''s denominators are relatively prime, meaning that the soonest it can overlap is when the first event has happened 59 times and the second event has happened 11 times, but we are back at 12 again, so it can''t happen.)

Brendan

very interesting math problem. Let''s we check with an equation.

It is easy to say that the hour and minute hands of the clock overlap 11 times every circle (half day, I mean). So we can check where the second hand is when every time the hour and minute hands overlap to check if the three hands overlap.

1. All three hands overlap at 12 o''clock.

2. After 12 o''clock, hour hand and minute hand should overlap between 1 and 2 o''clock. So let''s assume that from 1 o''clock to the time the two hands first overlap, the hour hand rotates x rad and minute hand rotates y rad

y = 2*PI / 12 + x;
y = 12*x;

so x = (2*PI/12) / 11;

the hour and minute hands overlap at 2*PI/11 rad. The time is
1/11 hour pass one o''clock. (note 2*PI/12 is one hour''s rotation). At this time, the second hand should be at (3600 / 11) % 60 rad. You can check it is not overlap with the other two hands.

3. Using the same way, you can check all the possibilities.

kidd

Basically, just calculate it like Punty50 did:

quote:

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Original post by Punty50
...My other way of thinking about this problem was that this event happens 11 times in 12 hours, so you divide out and get the same answer...

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11 times in 12 hours, the hour and minute hand overlap. Check to see where the second hand is at these times (since these are the only times that the other 2 overlap), and you''ll have your answer.

Jason Doucette
www.jasondoucette.com

Right-
It wont happen because the two events (hour and minute, minute and second) will never occur at the same time because the one happens 12/11th of an hour (720/11) minutes, and the other happens 60/59th minutes. They don''t share a common demoninator, which means that the only time they will overlap is after they both reach the original position.

Brendan