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# Inverting Clocks!

### #1

Posted 20 December 2007 - 01:15 PM

Note:

Invert = replace each one with the other

Possible time = time that can be seen on a clock

Ex: if it's 6 o'clock, the hour hand is on 6 and the minute hand is on 12. Now if we invert the hands, we get the hour hand on 12 and the minute hand on 6. This is not a "possible" time because at 12:30 the hour hand is midway between 12 and 1.

Again, there is a complicated way to solve it and another much simpler way!

### #2

Posted 21 December 2007 - 01:47 AM

### #3

Posted 21 December 2007 - 02:01 AM

How many times in a day can we invert the hour and minute hands, and have the clock still showing a "possible" time?

Note:

Invert = replace each one with the other

Possible time = time that can be seen on a clock

Ex: if it's 6 o'clock, the hour hand is on 6 and the minute hand is on 12. Now if we invert the hands, we get the hour hand on 12 and the minute hand on 6. This is not a "possible" time because at 12:30 the hour hand is midway between 12 and 1.

Again, there is a complicated way to solve it and another much simpler way!

### #4

Posted 21 December 2007 - 02:02 AM

### #5

Posted 21 December 2007 - 01:36 PM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #6

Posted 21 December 2007 - 06:57 PM

### #7

Posted 23 December 2007 - 01:06 AM

Interested to knowHow many times in a day can we invert the hour and minute hands, and have the clock still showing a "possible" time?

... there is a complicated way to solve it and another much simpler way!

**roolstar**'s approach.

Care to share, roolstar?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #8

Posted 24 December 2007 - 12:57 PM

Consider the hour between noon and one o'clock. The hour hand starts at the zero-minute position and moves as far as the five-minute position. Including twelve o'clock itself, when the hour and minute hand coincide, the minute hand will pass a "bit past the hour" point for one o'clock, two o'clock, and so on through eleven o'clock, for a total of twelve times in the first hour when the hour and minute hands can be swapped and still yield a possible time. For the second hour, when the hour hand is between five and ten minutes past the hour, same story -- twelve possible positions. The is true completely through the ten o'clock hour, so the first eleven hours each give twelve swappable positions.

I'm not sure your reasonning is correct unless I'm missing something here. it looks to me like you said: "the minute hand and the hour hand will coincide TWELVE times EVERY hour! And that, of course is simply not true. In fact the minute and hour hands will only coincide ONCE every hour. In fact you yourself said "a bit past the hour", this means we are already in the 1 o'clock.

Now bonanova:

Imagine a clock hand that moves at 1/5 the speed of a normal second hand. Let's call that imaginary hand the fifth hand.

Beginning at midnight, when the fifth hand will have made its first revolution, five minutes will have elapsed,

and the minute hand will have moved from 12 to 1. Thus the fifth hand and the minute hand behave

together just as the minute and hour hands do: At each moment in time, the fifth hand occupies the

position a minute hand would occupy if the hour hand were at the position of the minute hand.

Great reasonning. We simply created a "fifth hand" that goes 12 times faster than the minute hand; exactly like the minute hand goes 12 times faster than the hour hand. That was my "much simpler approach" too. Simple doesn't always mean easy to grasp though.

Another approach however is Mechanical and based on Geometry not on logic alone. I prefer it because we can easilly calculate each possible position with a simple 2 equations system.

Can you figure out what I mean?

### #9

Posted 25 December 2007 - 10:09 AM

Another approach however is Mechanical and based on Geometry not on logic alone. I prefer it because we can easilly calculate each possible position with a simple 2 equations system.

Can you figure out what I mean?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #10

Posted 25 December 2007 - 10:17 PM

As for spoxjox:

Consider the hour between noon and one o'clock. The hour hand starts at the zero-minute position and moves as far as the five-minute position. Including twelve o'clock itself, when the hour and minute hand coincide, the minute hand will pass a "bit past the hour" point for one o'clock, two o'clock, and so on through eleven o'clock, for a total of twelve times in the first hour when the hour and minute hands can be swapped and still yield a possible time. For the second hour, when the hour hand is between five and ten minutes past the hour, same story -- twelve possible positions. The is true completely through the ten o'clock hour, so the first eleven hours each give twelve swappable positions.

I'm not sure your reasonning is correct unless I'm missing something here. it looks to me like you said: "the minute hand and the hour hand will coincide TWELVE times EVERY hour! And that, of course is simply not true. In fact the minute and hour hands will only coincide ONCE every hour. In fact you yourself said "a bit past the hour", this means we are already in the 1 o'clock.

Yes, you are correct; you misunderstood me. As the parts in blue indicate, I am referring to the idea that at about 12:05:30, 12:11:00, 12:16:30, etc. the minute hand will reach a point where, if it were swapped with the hour hand, the clock would read a possible time. Thus, there are twelve such positions for each of eleven hours, and eleven such positions for the twelfth hour, giving 143 total for each twelve-hour period.

Merry Christmas!

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