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Math brain teaser with clock hands

tim-from-pa

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This is a teaser but it does have an absolute answer. I think most here can get the right answer with a little careful thought. After we get a few answers, I'll state the significance of this teaser regarding something similar that's rather educational. (And yes, I won't let people hanging but will give the correct answer eventually)

Suppose I have a clock with a smooth running hour and minute hand? It's 12:00 midnight and they are together. At what time (to the nearest second) is it when the minute hand catches up with the hour hand again?
 
What kind of clock? 24hr clock? 12hr clock? :dunno :shrug :confused
 
Cool, I didn't know the pointers in a clock are called "hands"! :D

The minute hand makes 60/360 = 6 degrees of the circle each minute.
The hour hand is slower, it needs 1/(12*60) = 0.5 of a degree each minute.

Both starting at position 0:00 the minute hand goes ahead, makes a whole turn in 60 min. During that time the hour hand has gone 30 degrees (it's on the 1 o'clock position now).
When the minute hand reaches 1 o'clock (5 minutes later) the hour hand has advanced by 2 and a half degrees. It's on 32.5 deg now. It's gonna take the minute hand 25 seconds to catch up. But the minute hand has moved on even in that short time. Will this turn into an infinitesimal calculus task? :o Will the two hands ever meet again? Maybe their tragical fate is to never see each other again, because no matter how small the angle the hour hand has moved, it will take the minute hand an amount of time > 0 to catch up, during which the hour hand can escape. They will chase each other for all eternity and never get together. :crying

Practically we know they will be on the same position at about 01:05:25.
 
Actually my digital 12hr clock is perfect and accurate. Check at 1:00 dot
 
Cool, I didn't know the pointers in a clock are called "hands"! :D

The minute hand makes 60/360 = 6 degrees of the circle each minute.
The hour hand is slower, it needs 1/(12*60) = 0.5 of a degree each minute.

Both starting at position 0:00 the minute hand goes ahead, makes a whole turn in 60 min. During that time the hour hand has gone 30 degrees (it's on the 1 o'clock position now).
When the minute hand reaches 1 o'clock (5 minutes later) the hour hand has advanced by 2 and a half degrees. It's on 32.5 deg now. It's gonna take the minute hand 25 seconds to catch up. But the minute hand has moved on even in that short time. Will this turn into an infinitesimal calculus task? :o Will the two hands ever meet again? Maybe their tragical fate is to never see each other again, because no matter how small the angle the hour hand has moved, it will take the minute hand an amount of time > 0 to catch up, during which the hour hand can escape. They will chase each other for all eternity and never get together. :crying

Practically we know they will be on the same position at about 01:05:25.

Sometimes we think alike, that's how I use to do it. Hint: approach velocity. I'll let some others answer for awhile yet.
 
This is a teaser but it does have an absolute answer. I think most here can get the right answer with a little careful thought. After we get a few answers, I'll state the significance of this teaser regarding something similar that's rather educational. (And yes, I won't let people hanging but will give the correct answer eventually)

Suppose I have a clock with a smooth running hour and minute hand? It's 12:00 midnight and they are together. At what time (to the nearest second) is it when the minute hand catches up with the hour hand again?
1:05.21 AM
 
Last edited by a moderator:
OK. I think this had enough time :lol for those who wanted to answer.

The time it would catch up to the nearest second would be (drum roll)

1:05:27 AM.

The minute hand passes the hour hand 11 times in 12 hours or every 12/11 hours. I'll let y'all carry out the math for that.

Another way to think of it is exactly the way Claudya was doing it : approach velocity. We can think of the minute hand as having 1 turn per 1 hour (1/1=1) and the hour hour 1 turn per 12 hours or 1/12 velocity. Thus the minute hand approaches the hour hand, (as if the hour hand was standing still) at the rate of 1-(1/12)=11/12 turns per hour, or that's 12 hours per 11 turns, the same fraction as above, and again we get 1:05:27.

Practical usage (scientific part):

When the planets orbit the sun, they go at different rates. The planet Venus orbits faster than earth completing an orbit in about 225 days compared to the earth's 365 days, sort of like one clock hand going faster than the other. When the planets relative to the sun are in the same configuration, say for example Venus between the earth and sun (also called transit if over the sun), is called the synodic period and it works out to be just over 19 months.

I like using the approach velocity method better because basically you just subtract the inverses and then invert your answer again, i.e. (1/225)-(1/365) then invert the answer. The reason is simply because unlike hour and minute hands, the orbits of planets are not a nice even multiple relative to the others.
 
It's amazing how often you find that principle of "something circling around something else" in nature and in artificial constructions. God must really love building orbital systems.

My idea that the minute hand might never reach hour hand was derived from that ancient greek paradox story of Achilles racing an old and slow turtle that he can never overtake because the turtle is slow, but it always gains some distance during the time Achilles needs to catch up. It's very interesting, because that paradox shows how mathematics can play tricks on your mind.
 
You're right. I redid the math. 1 hour, 27 point 2727272727 (ad infinitum) seconds. Don't how I missed that the first time around.
 
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