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In theory, this should be easy

  1. May 29, 2008 #1
    In theory, this should be easy....

    ... During the space of 12 hours, how many times will the big hand of a clock form a right angle with the little hand?
  2. jcsd
  3. May 29, 2008 #2
    23 times ?
  4. May 29, 2008 #3
    Close but no cigar.
  5. May 30, 2008 #4
    It depend upon when you start counting the 12 hours.
  6. May 30, 2008 #5
    59x2--each click on the clock can have a right hand right angle and left handed one...
  7. May 30, 2008 #6
    Yes. If you start counting at 3:00 or 9:00 for 12:00+epsilon hours, with epsilon positive as small as you want, it seems it makes 23 times. If epsilon is negative, that probably makes only 22. :smile:
  8. May 30, 2008 #7
    Does it have an hour, minute as well as second hand? Or only hours and minutes?
  9. May 30, 2008 #8
    OOps--forget previous answer...thinking 24 now...
  10. May 30, 2008 #9
    I was thinking second hand before
  11. May 30, 2008 #10
    Completely ignore the seconds hand, we are only considering the minutes and hours hand. For argument's sake, start at the current time as I post this here in my timezone, which is 1.05.

    But it seems to me that humanino has probably got it. :)
  12. May 30, 2008 #11
    Well, I got the first (3:00) yesterday before going to bed, and I stupidily was happy about myself, so I forgot the other one (9:00) :smile:
  13. May 31, 2008 #12
    There are others as well.
  14. May 31, 2008 #13
    Yes, so difficult the problem
  15. May 31, 2008 #14


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    I have used a "brute force" method, simply writing down the (approximate) times at which right-angles occur. Highlight the text below for the solution.

    Also ... somebody may want to check if I missed any :smile:

    There are 22 times where right-angles occur.

    Note: times are approximate.

    Last edited: May 31, 2008
  16. May 31, 2008 #15
    There is an analytic way to approach the problem--

    The position vector of the big hand is given by (where t=0 is noon)

    [tex]\mathbf{x_b}(t) = \sin \omega_b t \mathbf{i} + \cos \omega_b t \mathbf{j}[/tex]

    while the position vector of the little hand is given by

    [tex]\mathbf{x_l}(t) = \sin \omega_l t \mathbf{i} + \cos \omega_l t \mathbf{j}[/tex]

    and you require that they are perpendicular to each other so that

    [tex]\mathbf{x_b}(t)\cdot\mathbf{x_l}(t) = 0 \Rightarrow[/tex]

    [tex]\cos\omega_b t \cos\omega_l t + \sin \omega_b t \sin \omega_l t = 0 \Rightarrow [/tex]

    [tex]\cos (\omega_l - \omega_b)t = 0 \Rightarrow [/tex]

    [tex](\omega_l - \omega_b)t_n = \pi n \Rightarrow [/tex]

    [tex]t_n = \frac{\pi n}{\omega_l - \omega_b}[/tex]


    [tex]\omega_b = 2\pi/T_b, \omega_l = 2\pi/T_l[/tex] where the T's are the periods.

    So we have

    [tex]t_n = \frac{n}{2} \frac{T_b T_l}{T_b - T_l}[/tex]
    [tex]T_b = 12 h, T_l = 1 h[/tex]
  17. May 31, 2008 #16


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    Well its fairly easy to work out. angular velocity of each hand is simple.

    Angular velocity of minute hand: [itex] \omega_m=\frac{2\pi}{3600} [/itex]
    Angular velocity of hour hand: [itex] \omega_h = \frac{2\pi}{43200} [/itex]

    Goverened by the rotational dynamics equations:

    [tex] \theta = \theta_0 + \omega t[/tex]

    We want:

    [tex] \theta_m - \theta_h = (2n+1) \frac{\pi}{2} [/tex]

    and we get:

    [tex] t = \frac{(2n+1) \pi}{2(\omega_m - \omega_h)} [/tex]

    First couple of answers are (assuming you start from 12:00):


    EDIT: I was wondering why my tex had come out as vectors for a second.
  18. May 31, 2008 #17
    Oh yeah woops I made a rookie mistake-- should have been (2n+1)/2 \pi as Kurt has it and not \pi n. Duh!
  19. May 31, 2008 #18
    You have the right number. Well done.
    Last edited: May 31, 2008
  20. Jun 6, 2008 #19
    wow someone brought maths into it.
    i'd say for the first hour
    when the minute hand is 15mins past the hour hand
    for the last hour
    when the minute hand is 15mins befor the hour hand

    for the 10 hours in between
    at 15mins before and after the hour hand

    10*2 + 2*1 =22?
  21. Jun 12, 2008 #20
    Are we counting every second, minute or hour? LOL, J/K.
  22. Jun 16, 2008 #21


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    This is pretty easy stuff:

    The big hand makes 11 laps relative to the small hand. There are 2 times per lap that it forms a right angle, so the number is 2*11=22.
  23. Jun 17, 2008 #22
    What's with making the text invisible? It doesn't even work, the text is still visible.
  24. Jun 17, 2008 #23


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    To prevent spoilers to those who want to solve it themselves is the purpose. You have to use the correct colour however which is e3e3e3 I believe.

    Edit: I have seen other forums with some sort of system that blacks out the text of spoilers which you have to mouse over to reveal. I wonder if that could be applied to this forum.
  25. Jun 17, 2008 #24


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  26. Jun 17, 2008 #25


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    Comes out as e3e3e3 on photoshop for me. Both should work fine though.
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