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Puzzle: Exchangeable Clock Positions

  1. Dec 27, 2011 #1
    I was hoping some of the brilliant minds here might share their thoughts on the following puzzle:

    Definition. A valid clock position is exchangeable iff swapping the positions of the minute hand and the hour hand yields another valid clock position.

    Example. Midnight is an exchangeable clock position.

    Puzzle. How many exchangeable clock positions exist, and what are they?

    Partial Solution. We've found one special case already. By symmetry we can argue that there must be either 12 more exchangeable positions (one within each hour), or there must be infinitely many. The former seems intuitively more likely.

    Visually, a time of roughly 12:41 looks as though it may be exchangeable with a time near 8:03.5. And it seems like there ought to be 11 more of these for each hour.

    We can represent the minute hand as [tex] e^{i\frac{2\pi}{12}m}[/tex] where [tex]m[/tex] is between 0 and 60.

    And we can represent the hour hand as [tex] e^{i\frac{2\pi}{12}h}[/tex] where [tex]h[/tex] is between 0 and 12.

    Then as [tex]m[/tex] goes from 0 to 60 the hour hand, [tex]h[/tex], must go from 0 to 1. So we have
    [tex] h = \frac{m}{60}[/tex],
    if [tex]m[/tex] is the total number of minutes passed. But by this logic I found that the only exchangeable clock position is 12:00.

    I'm curious what everyone else thinks! Are there more exchangeable positions?
     
  2. jcsd
  3. Dec 27, 2011 #2

    Dick

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    No, no. The obvious solutions are that they can be obviously exchanged if the hour position equals the minute position. How many times does that happen? How in detail did you conclude there is only one??
     
    Last edited: Dec 27, 2011
  4. Dec 28, 2011 #3
    I may be misunderstanding you, but if the hour position equals the minute position then I think exchanging the hands will not yield a valid clock position unless the time is 12:00.

    Here's now I found that there's only one valid position. Let the number of minutes elapsed be denoted m. Then the hour position is h= m/60 (where the position on the clock is denoted 1 to 12). Now exchanging the hands means that we now have m' = m/60 and h' = m. Requiring that this new clock position be valid is the same as imposing the constraint h' = m'/60 which implies that h' = m = m'/60 = m/(60*60). Thus, m = m/(60*60), which implies that h = m = 0. So the only exchangeable clock position is 12:00.

    I would've guessed there would be more exchangeable positions, but I'm curious to see what you think. Perhaps there's another way to approach this, or my argument is flawed.
     
  5. Dec 28, 2011 #4

    Dick

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    That's bad notation. m='number of minutes' is not a clock position. If you want to exchange positions both of your values should represent clock positions. Try it this way. If h is the number of hours then the minute clock position m=h*12. For example, if h=(1/2) (30 minutes past 12) then m=(1/2)*12=6. So the minute hand is pointing at 6. That's the 'clock position'. And don't forget you'll need some kind of 'mod' function to account for the minute hand wrapping around.
     
  6. Dec 28, 2011 #5
    If you have the time 1:05 then both the minute hand and hour hand are pointing at the '1' on a typical clock. If you exchange both of the hands they are still pointing at the '1' on the clock. This already proves your initial conclusion wrong. Don't automatically conclude that your math is right, think about the question first and try and see if your solution is reasonable, it clearly isn't.
     
  7. Dec 28, 2011 #6

    SammyS

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    [STRIKE]I think you are correct, and your argument is valid.[/STRIKE] See Clever-Name's post.

    It's easy to get fooled by being led into thinking a particular way.


    Another way to look at this is as follows: (Unlike your argument, it's not a general argument, just a specific example.)
    Let's look at two times which appear to have exchangeable clock positions; some time near 12:55 (5 minutes before 1 o'clock) and some other time, near 11:05.

    A necessary condition for a time to have exchangeable clock positions is that the angle the minute hand makes with the vertical has the same measure as the angle the hour hand make with the vertical.

    At exactly 12:55, the minute hand is 30° from vertical, the hour hand is 29.5° from vertical. Less than 1/2 a minute later the angles will have equal measure.

    At exactly 11:05, the minute hand is 30° from vertical, the hour hand is 30.5° from vertical. Less than 1/2 a minute earlier the angles had equal measure.​

    So, there is no time close to 11:05 or 1:55, at which the clock positions are exchangeable.

    Added in Edit (after reading Clever-Name's post immediately above):

    I have amended this post (using the strike-out feature and displaying new material in red).
     
    Last edited: Dec 28, 2011
  8. Dec 28, 2011 #7

    Dick

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    Sure. But the time near 1:05 when the hands overlap isn't EXACTLY at 1:05, right? The hands can't both point exactly at 1.
     
  9. Dec 28, 2011 #8

    Dick

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    If you are agreeing 12:00 is the only solution, that's quite wrong. At time 12/11 hours (the ~1:05 time Clever-Name as referring to) the hands point in the same direction. And there are lots of other solutions where the hands don't point in the same direction.
     
  10. Dec 28, 2011 #9

    SammyS

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    Right, but at 1:05, the minute has not yet overtaken the hour hand. By 1:06, the minutes hand will have overtaken the hour hand. If the both move smoothly, they both will be at the same position at some time between 1:05 and 1:06.

    My apologies, if my previous post misled anyone, during the time I was editing it !
     
  11. Dec 28, 2011 #10

    SammyS

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    5/11 of a minute after 1:05, the hands will be at the same location.
     
  12. Dec 28, 2011 #11

    Dick

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    Right. That's 12/11 hours after 12:00. The hands are at the same location again at 2*12/11 hours after 12:00. Etc. I think these are the sort of 'obvious' solutions the 'Partial Solution' was inaccurately talking about in the OP. Like I said before, there are other solutions.
     
  13. Dec 28, 2011 #12
    Well of course, it wont be EXACTLY at the 1, it'll be a little displaced towards the 2, I was merely using that as an approximation to motivate my point in the rest of my post
     
  14. Dec 28, 2011 #13
    Ah yes I misread this. Exchanging has no effect when the positions are equal.

    But like Clever-Name and SammyS have also suggested, there appear to be many other exchangeable positions where the hands aren't equal.

    These are exactly the positions I was referring to! I believe based on the 12-fold symmetry of the clock, we should find 12 of these positions. I just don't know how to find them.
     
  15. Dec 28, 2011 #14

    Dick

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    You were really pretty close in post 3. I gave you some suggestions in post 4 on how to fix it up. Did you try thinking about those?
     
  16. Dec 28, 2011 #15
    Ah yeah -- now it clicks! Thank you! What you said makes perfect sense, since as the hour hand moves from say h to (h+1), the minute hand makes a full revolution. I'm just about to work out the solutions...
     
  17. Dec 28, 2011 #16
    Imagine two clocks. One is normal, but the other one has an unusual hour "hand", which I'll call a dodecapus - it is twelve hands pointing at all the hours. Now we can drive the normal-looking clock in an unusual way - by pushing the hour hand. As we do this, the normal minute hand moves twelve times as fast, of course. Meanwhile we'll drive the unusual clock by pushing the minute hand, keeping it exactly aligned with the hour hand on the other clock. As we do this, we can see the normal minute hand will align with each hand of the dodecapus in turn - generating an exchangeable position. The normal minute hand circles the clockface twelve time, while the dodecapus has moved on one-twelth of the circuit. This gives a method to get to the count... or at least a good guess at it.
     
  18. Dec 29, 2011 #17
    Here are some steps toward a solution.

    Let h(t) be the angle swept out by the hour hand, in degrees, at time t minutes after starting the clock, and m(t) be the angle swept out by the minute hand, so
    [tex]h(t) = t / 2[/tex] and
    [tex]m(t) = 6t[/tex]
    Then times [itex]t_1[/itex] and [itex]t_2[/itex] are exchangeable if
    [tex]h(t_1) = m(t_2)[/tex] and
    [tex]m(t_1) = h(t_2)[/tex] as angles.

    Remember that two angles are "equal" if they differ by an integer multiple of 360 degrees.
     
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