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Hands On A Clock (Posted on 2004-07-05) Difficulty: 3 of 5
Do the three hands on an analog clock (hours, minutes, seconds) ever divide the face of the clock into three equal segments, i.e. 120 degrees between each hand?

No Solution Yet Submitted by SilverKnight    
Rating: 3.5000 (10 votes)

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Solution Brute Force (i.e., exhaustive search) | Comment 1 of 23

As the relative positions of the hour and minute hand repeat in a cycle of 12/11 hours, and the hour and minute hands are together at 12:00, they will meet each other at increments of 120 degrees (including zero) every 12/33 of an hour.  The following program lists all such times, with the angular positions of the hour hand, minute hand and second hand, and the differences between each pair of hands, assuming an ideal clock with smoothly moving hands and no play in the gears:

DEFDBL A-Z
FOR t = 0 TO 12 STEP 12 / 33
  h = INT(t)
  m = (t - h) * 60
  s = (m - INT(m)) * 60
  hh = 360 * t / 12
  mh = 360 * m / 60
  sh = 360 * s / 60
  PRINT USING "##:##.##"; h; m;
  PRINT USING "#####.####"; hh; mh; sh; hh - mh; mh - sh; sh - hh
NEXT t

The resulting table:

 0: 0.00    0.0000    0.0000    0.0000    0.0000    0.0000    0.0000
 0:21.82   10.9091  130.9091  294.5455 -120.0000 -163.6364  283.6364
 0:43.64   21.8182  261.8182  229.0909 -240.0000   32.7273  207.2727
 1: 5.45   32.7273   32.7273  163.6364    0.0000 -130.9091  130.9091
 1:27.27   43.6364  163.6364   98.1818 -120.0000   65.4545   54.5455
 1:49.09   54.5455  294.5455   32.7273 -240.0000  261.8182  -21.8182
 2:10.91   65.4545   65.4545  327.2727   -0.0000 -261.8182  261.8182
 2:32.73   76.3636  196.3636  261.8182 -120.0000  -65.4545  185.4545
 2:54.55   87.2727  327.2727  196.3636 -240.0000  130.9091  109.0909
 3:16.36   98.1818   98.1818  130.9091   -0.0000  -32.7273   32.7273
 3:38.18  109.0909  229.0909   65.4545 -120.0000  163.6364  -43.6364
 4: 0.00  120.0000    0.0000    0.0000  120.0000   -0.0000 -120.0000
 4:21.82  130.9091  130.9091  294.5455   -0.0000 -163.6364  163.6364
 4:43.64  141.8182  261.8182  229.0909 -120.0000   32.7273   87.2727
 5: 5.45  152.7273   32.7273  163.6364  120.0000 -130.9091   10.9091
 5:27.27  163.6364  163.6364   98.1818    0.0000   65.4545  -65.4545
 5:49.09  174.5455  294.5455   32.7273 -120.0000  261.8182 -141.8182
 6:10.91  185.4545   65.4545  327.2727  120.0000 -261.8182  141.8182
 6:32.73  196.3636  196.3636  261.8182    0.0000  -65.4545   65.4545
 6:54.55  207.2727  327.2727  196.3636 -120.0000  130.9091  -10.9091
 7:16.36  218.1818   98.1818  130.9091  120.0000  -32.7273  -87.2727
 7:38.18  229.0909  229.0909   65.4545    0.0000  163.6364 -163.6364
 7:60.00  240.0000  360.0000  360.0000 -120.0000    0.0000  120.0000
 8:21.82  250.9091  130.9091  294.5455  120.0000 -163.6364   43.6364
 8:43.64  261.8182  261.8182  229.0909    0.0000   32.7273  -32.7273
 9: 5.45  272.7273   32.7273  163.6364  240.0000 -130.9091 -109.0909
 9:27.27  283.6364  163.6364   98.1818  120.0000   65.4545 -185.4545
 9:49.09  294.5455  294.5455   32.7273    0.0000  261.8182 -261.8182
10:10.91  305.4545   65.4545  327.2727  240.0000 -261.8182   21.8182
10:32.73  316.3636  196.3636  261.8182  120.0000  -65.4545  -54.5455
10:54.55  327.2727  327.2727  196.3636    0.0000  130.9091 -130.9091
11:16.36  338.1818   98.1818  130.9091  240.0000  -32.7273 -207.2727
11:38.18  349.0909  229.0909   65.4545  120.0000  163.6364 -283.6364
11:60.00  360.0000  360.0000  360.0000    0.0000    0.0000   -0.0000

shows no row in which all of the last three columns show +/-120 or +/- 240, so the answer is no.

(Please forgive the 11:60 as a reference to 12:00, etc. as the internal representation of numbers as binary leads to truncation so that the hours were just short of 12 and the integer part fell to 11. Likewise 7:60 for 8:00.)


  Posted by Charlie on 2004-07-05 11:00:30
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