Clock Puzzle: A regular analogue clock...?

Clearly, when the hands coincide, it works.

That occurs once per hour,

around 3:16, 4:21, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55, and 11:60 (better known as 12:00).

(9 times).

Slightly more precise (1 second resolution), but not quite the times are

3:16:22

4:21:49

5:27:16

6:32:44

7:38:11

8:43:38

9:49:05

10:54:33

It's quite tricky to calculate if other times work as well.

Between 3 and 12, the hour hand is ... between 3 and 12 (of course)

which means we are only interested in minutes between 15 and 60 (or 0).

If we make the 12 equal to 0 degrees, then

1 2 3 . . . . 9 10 11 represent

30 60 90 ... 270 300 330 degrees.

The minute hand advances at 6° per minute,

and the hour hand advances at 1/2° per minute.

At m minutes past h hour,

the minute hand is at 6m °

the hour hand is at (30h + m/2) °

Consider the 3:00 to 3:59 hour.

During that time, the hour hand is between 3 and 4,

which if it were the minute hand would represent

15 to 20 minutes past the hour.

If the minute hand is between 4 and 5,

then swapping them represents times between 4:15 and 4:20.

Let m = minutes past 4:00.

h degrees = 120 + m/2

m degrees = 6m (m between 15 and 20, m degrees between 90 and 120).

If we swap them, we have a time between 3:20 and 3:25.

And we want to find p such that

120 + m/2 = 6p and

6m = 90 + p/2

240 + m = 12p

144m = 2160 + 12p

240 + m = 144m - 2160

143m = 2400

m = 16.78

At 4:16.78 minute hand is at 100.68 degrees,

and the hour hand is at 128.39 degrees.

Reversing those gives a time of 3:21.4 which also corresponds to those degrees.

Then rather than solve such an equation for every 5-minute interval,

I resorted to a program, which prints out times when it is "close".

Here is the output:

note on the left of the -> is hh:mm:ss

and on the right it's hh:mm.fraction of a minute

There are 45 on this list.

I computed them using 1 second intervals.

It may be that for exact accuracy you need sub-second

computations ... and I also computed within a tolerance

of 0.05 degrees.

It might be the case that they don't work exactly.

1. 3:16:22 -> 3:16.36

2. 3:21:24 -> 4:16.78

3. 3:26:26 -> 5:17.20

4. 3:31:28 -> 6:17.62

5. 3:36:30 -> 7:18.04

6. 3:41:32 -> 8:18.46

7. 3:46:34 -> 9:18.88

8. 3:51:36 -> 10:19.30

9. 3:51:37 -> 10:19.30

10. 3:56:39 -> 11:19.72

11. 4:21:49 -> 4:21.82

12. 4:26:51 -> 5:22.24

13. 4:31:53 -> 6:22.66

14. 4:36:55 -> 7:23.08

15. 4:41:57 -> 8:23.50

16. 4:47: 0 -> 9:23.92

17. 4:52: 2 -> 10:24.34

18. 4:57: 4 -> 11:24.76

19. 5:27:16 -> 5:27.27

20. 5:32:18 -> 6:27.69

21. 5:37:21 -> 7:28.11

22. 5:42:23 -> 8:28.53

23. 5:47:25 -> 9:28.95

24. 5:52:27 -> 10:29.37

25. 5:57:29 -> 11:29.79

26. 6:32:44 -> 6:32.73

27. 6:37:46 -> 7:33.15

28. 6:42:48 -> 8:33.57

29. 6:47:50 -> 9:33.99

30. 6:52:52 -> 10:34.41

31. 6:57:54 -> 11:34.82

32. 7:38:11 -> 7:38.18

33. 7:43:13 -> 8:38.60

34. 7:48:15 -> 9:39.02

35. 7:53:17 -> 10:39.44

36. 7:58:19 -> 11:39.86

37. 8:43:38 -> 8:43.64

38. 8:48:40 -> 9:44.06

39. 8:53:42 -> 10:44.48

40. 8:58:44 -> 11:44.89

41. 9:49: 5 -> 9:49.09

42. 9:54: 8 -> 10:49.51

43. 9:59:10 -> 11:49.93

44. 10:54:33 -> 10:54.55

45. 10:59:35 -> 11:54.97

You can also reverse any A -> B to get B -> A.

Since the hour hand rotates 12 times as fast as the minute hand, supposing A and B are the angles made by hour and minute hand resp. to 12 o-clock,

12A = B + 2nXpi where n is any integer

Since the same relationship should hold with the two hands swapped,

12B = A + 2kXpi

We get A-B = 2(n-k)pi/13, and

A+B = 2(n+k)pi/11

Since n and k are integers, so are n-k and n+k, and we have

A-B=2mpi/13

A+B=2qpi/11, with the stipulation that m and q be both even or both odd.

Assume without loss of generality that A>=B

Since A and B must be between 3 and 12 pm, each of them lies between pi/2 and 2pi. Hence we have

0 <= A-B <= 3pi/2

and, pi <= A+B <= 4pi

Thus m can be 0, 1, 2, ...,9 and q can be 6, 7, 8, ..,22

There are 5 odd m values, and 8 odd q values, which is 40 pairs of odd (m,q)

There are 5 even m values, and 9 even q values, which is 45 pairs of even (m,q)

(Remember that m and q cannot be of different parities)

So we have a total of 85 solutions, for now.

The solutions are A=(m/13 + q/11)pi and B= (q/11 - m/13)pi where (m,q) are each of the pairs mentioned above. Since A and B must be between 3pi/2 and 2pi, this further reduces the possible pairs. I suggest searching systematically or using a computer program with a simple "if" loop to do it for you.