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## Question

12 o’clock is the only time when all three hands on the clock coincide exactly. (As already has been established by similar problem(s) on the BrainDen.)

I estimate, in a 12-hour period, any two hands on a clock coincide approximately 1438 times. (Did I get that right?)

Other than 12 o’clock, at what time(s) does the smallest angle between one of the clock’s hands and the other two that have coincided occur?

Assume the clock hands motion is continuous, smooth and even.

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• 0 ummmm..

1 second before 12 o clock, or 11:59:59?

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• 0 I'm thinking 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, 10:50 and 11:55

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• 0 ummmm..

1 second before 12 o clock, or 11:59:59?

Once the second hand moves even a tiny distance, the minute clock will have moved. So the shortest distance would be when the second hand sweeps around and is coincident with the hour hand near 12 at just over 1 minute past 12, and when it sweeps over the minute hand at just over 1 minute before 12.

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• 0 I think Like this

we know that at 5minut 27.2727... minute past 1 o' **** minute and hour hand overlap. now when the second hand comes round to overlap the hour hand the minute hand would have swept an anlge of time --- 60-27.27= 32.727272 + around 5 = 37.7 seconds ---

also at, 53 minutes, 27.27 seconds past 11 o' clock, minute and hour hand will overlap. Now if second hands moves forward to overlap hourhand then, after it has completed the task, the minute hand will have swept and angle of time ---- about 53-27.2727 = 25.723 seconds--- less than the above case.

I don't think this is minimum. So, keep on checking for all 11 minute hour, overlap cases and you get the answer.

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• 0 In my previous post its 53minutes 27.27 seconds past 10' 0 clock, not 11'o clock. Sorry for the mistake.

Ok, I think now got it

at 43 minutes 38.1818 seconds past 8 o' clock minute and hour hand overlap. after about 4.1818 seconds the second hand overlaps with hour hand. At this time the minute hand has swept an angle of only --- 4.1818 seconds time--- with the overlaped hour-second hand. This, I think, is the minimum possible such angle.

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• 0 ...what if U don't have a face clock?...I'm trying 2 figure this out but all my clocks are digital...

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• 0 ...what if U don't have a face clock?...I'm trying 2 figure this out but all my clocks are digital...

What is a face clock, dear?

And, You don't need to have any clock. Its a pen and paper work.

Edited by the-genius

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• 0
In my previous post its 53minutes 27.27 seconds past 10' 0 clock, not 11'o clock. Sorry for the mistake.

Ok, I think now got it

at 43 minutes 38.1818 seconds past 8 o' clock minute and hour hand overlap. after about 4.1818 seconds the second hand overlaps with hour hand. At this time the minute hand has swept an angle of only --- 4.1818 seconds time--- with the overlaped hour-second hand. This, I think, is the minimum possible such angle.

That looks like the right answer, one of them, anyway. (They come in pairs because of the symmetry.) Although the actual calculations are still a bit off. For the sake of completeness here is the entire solution.

First, let’s figure out which two hands must coincide.

If the minute and the second hand coincided and the hour hand was slightly ahead, then moving the second hand to catch up with the hour would make a smaller angle with the minute hand.

Suppose, the hour and the minute hand coincided, then moving the second hand (forward or backward, whichever is closer) to coincide with the hour hand would move the minute hand away by a 60 times smaller distance apart than the second hand was.

Thus we can always improve the minute/hour hand coincidence and the minute/second hand coincidence by the nearest time when the hour and the second hand coincide.

It also stands the reason that for the minimal angle, the second/hour hand coincidence must be the nearest to a time when the minute and the hour hands coincided. Since, other than 12 o’clock, there are only 10 cases like that, we are relieved from checking all 1438 cases.

Now we must do some arithmetic.

As t-g pointed out, at the time when the hour and the minute hand coincide and the second hand is closest -- moving the second hand (forward or backward, whichever is closer) to coincide with the hour hand will give us the solution.

The minute hand catches up with the hour hand every 12/11 hours, or 1 and 1/11 hour.

For the purpose of angle, 12 o'clock is a good natural reference point. So let's discard the whole number of hours and use only the fractional part. Thus the first meeting of the hour and second hand occurs on 1/11th fraction of an hour (or the clock circle), the second -- on 2/11, third -- on 3/11, and so on.

Now let's check the second hand movement. In 1/11 of an hour it has traveled 60/11 of the circle, or 5 and 5/11. Again, we are only interested in the fractional part. So after 2/11 of an hour, the second hand is at 10/11 of the circle, after 3/11 hour -- at 4/11, and so on (using modular division).

So here are all 10 pairs of relative positions, top row the part of the circle where the minute and hour hands are, bottom row -- the second hand:

1/11 . . . 2/11 . . 3/11 . . 4/11 . . 5/11 . . 6/11 . . 7/11 . . 8/11 . . 9/11 . . 10/11

5/11 . . 10/11 . . 4/11 . . 9/11 . . 3/11 . . 8/11 . . 2/11 . . 7/11 . . 1/11 . . 6/11

As we can see, the second hand is closest to the hour/minute hand on their 3rd and 8th meetings. Then it is only 1/11th of the circle away.

The second hand catches up to the hour hand at the rate of 3595 steps per hour. So it would traverse 1/11 of the circle (or 60/11 steps) in 60/11/3595 hours. That subtracted from the 3 and 3/11 hours yields the approximate time of 3:16:16.356. (To obtain another optimal time, just subtract that from 12.) The minute hand runs away from the hour hand at the rate of 55 steps per hour. So when the second and the hour hands are together it would be 55*60/11/3595 = 0.0834 steps (minutes) away. A minute on a clock equals to an angle of 6o, that makes the angle between the clock hands approximately 0.5o.

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• 0 What is a face clock, dear?

And, You don't need to have any clock. Its a pen and paper work.

...face clock: 1 with hands...there's a bigger name that I cannot remember right now...

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• 0 That looks like the right answer, one of them, anyway. (They come in pairs because of the symmetry.) Although the actual calculations are still a bit off. For the sake of completeness here is the entire solution.

First, let’s figure out which two hands must coincide.

If the minute and the second hand coincided and the hour hand was slightly ahead, then moving the second hand to catch up with the hour would make a smaller angle with the minute hand.

Suppose, the hour and the minute hand coincided, then moving the second hand (forward or backward, whichever is closer) to coincide with the hour hand would move the minute hand away by a 60 times smaller distance apart than the second hand was.

Thus we can always improve the minute/hour hand coincidence and the minute/second hand coincidence by the nearest time when the hour and the second hand coincide.

It also stands the reason that for the minimal angle, the second/hour hand coincidence must be the nearest to a time when the minute and the hour hands coincided. Since, other than 12 o’clock, there are only 10 cases like that, we are relieved from checking all 1438 cases.

Now we must do some arithmetic.

As t-g pointed out, at the time when the hour and the minute hand coincide and the second hand is closest -- moving the second hand (forward or backward, whichever is closer) to coincide with the hour hand will give us the solution.

The minute hand catches up with the hour hand every 12/11 hours, or 1 and 1/11 hour.

For the purpose of angle, 12 o'clock is a good natural reference point. So let's discard the whole number of hours and use only the fractional part. Thus the first meeting of the hour and second hand occurs on 1/11th fraction of an hour (or the clock circle), the second -- on 2/11, third -- on 3/11, and so on.

Now let's check the second hand movement. In 1/11 of an hour it has traveled 60/11 of the circle, or 5 and 5/11. Again, we are only interested in the fractional part. So after 2/11 of an hour, the second hand is at 10/11 of the circle, after 3/11 hour -- at 4/11, and so on (using modular division).

So here are all 10 pairs of relative positions, top row the part of the circle where the minute and hour hands are, bottom row -- the second hand:

1/11 . . . 2/11 . . 3/11 . . 4/11 . . 5/11 . . 6/11 . . 7/11 . . 8/11 . . 9/11 . . 10/11

5/11 . . 10/11 . . 4/11 . . 9/11 . . 3/11 . . 8/11 . . 2/11 . . 7/11 . . 1/11 . . 6/11

As we can see, the second hand is closest to the hour/minute hand on their 3rd and 8th meetings. Then it is only 1/11th of the circle away.

The second hand catches up to the hour hand at the rate of 3595 steps per hour. So it would traverse 1/11 of the circle (or 60/11 steps) in 60/11/3595 hours. That subtracted from the 3 and 3/11 hours yields the approximate time of 3:16:16.356. (To obtain another optimal time, just subtract that from 12.) The minute hand runs away from the hour hand at the rate of 55 steps per hour. So when the second and the hour hands are together it would be 55*60/11/3595 = 0.0834 steps (minutes) away. A minute on a clock equals to an angle of 6o, that makes the angle between the clock hands approximately 0.5o.

Prime, you explained it well. But look here, YOu said

""""""""""""""""

1/11 . . . 2/11 . . 3/11 . . 4/11 . . 5/11 . . 6/11 . . 7/11 . . 8/11 . . 9/11 . . 10/11

5/11 . . 10/11 . . 4/11 . . 9/11 . . 3/11 . . 8/11 . . 2/11 . . 7/11 . . 1/11 . . 6/11

As we can see, the second hand is closest to the hour/minute hand on their 3rd and 8th meetings. Then it is only 1/11th of the circle away.

The second hand catches up to the hour hand at the rate of 3595 steps per hour.------ ------So it would traverse 1/11 of the circle (or 60/11 steps) in 60/11/3595 hours

""""""""""""""""

By the time the second hand travels 1/11 th of the circle the hour hand will also have moved. So, really the second hand don't cathes hour hand after 1/11, but very slightly more. Since you are going for a precise math, I think this should also be taken into account.

Taking this into account, I got the exact time when the angle is minimum to be-------- 8h:43m: (43+53/7909)s

Edited by the-genius

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• 0
Prime, you explained it well. But look here, YOu said

""""""""""""""""

1/11 . . . 2/11 . . 3/11 . . 4/11 . . 5/11 . . 6/11 . . 7/11 . . 8/11 . . 9/11 . . 10/11

5/11 . . 10/11 . . 4/11 . . 9/11 . . 3/11 . . 8/11 . . 2/11 . . 7/11 . . 1/11 . . 6/11

As we can see, the second hand is closest to the hour/minute hand on their 3rd and 8th meetings. Then it is only 1/11th of the circle away.

The second hand catches up to the hour hand at the rate of 3595 steps per hour.------ ------So it would traverse 1/11 of the circle (or 60/11 steps) in 60/11/3595 hours

""""""""""""""""

By the time the second hand travels 1/11 th of the circle the hour hand will also have moved. So, really the second hand don't cathes hour hand after 1/11, but very slightly more. Since you are going for a precise math, I think this should also be taken into account.

Taking this into account, I got the exact time when the angle is minimum to be-------- 8h:43m: (43+53/7909)s

I stand by my calculations. I said the second hand catches up to the hour hand at the rate of 3595 steps per hour. That already takes into the account the movement of the hour hand. The second hand moves at the speed of 3600 steps (minute divisions) per hour, the hour hand moves at the speed of 5 steps per hour. 3600-5=3595 is their relative speed. So if the second and the hour hands are 60//11 steps apart, it would take exactly 60/11/3595 hours for the second hand to catch up. I also used relative speed of the minute to the hour hand when calculating how far the minute hand would run away in that time.

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