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A previous clock puzzle made me think of this one:

At what times of the day are the hour hand and minute hand of an analog clock exactly at 90 degrees?

3 o'clock

9 o'clock

the two most obvious ones...

Edited by Romulus064
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It happens every 32:43 minutes starting at 12:16:22. So, exact times are:

12:16:22

12:49:05

1:21:49

1:54:33

2:27:16

3:00:00

3:32:44

4:05:27

4:38:11

5:10:55

5:43:38

6:16:22

6:49:05

7:21:49

7:54:33

8:27:16

9:00:00

9:32:44

10:05:27

10:38:11

11:10:55

11:43:38

Nice work! Could you show us the method you used to arrive at the 32:43 period?

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Nice work! Could you show us the method you used to arrive at the 32:43 period?

For period:

Assume t is any of times that provides 900 angle.

The next 900 angled time after this, the minute arm of the clock will have traveled m minutes = m.600 on the clock, the short arm will have traveled m.600/12 on the clock.

The statement:

m.600 - m.600/12 = travel difference.

If beginning angle is 900 that next angle will be 2700.

so the travel difference is 1800

then

m.600 - m.600/12 = 1800

m=36/11 = 32.72 minutes= 32:43

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It happens 22 times in 12 hours.

The spacing is 12/22 hours = 32 min 43.63636363... seconds.

30 * (1+1/12+1/12^2+1/12^3,...)

30 * 12/(1-1/12)

32.7272....minutes

Edited by G Threat
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