# ReTurn Time - 2

1. Apr 20, 2004

### hemmul

Here, we found the number of Arcon's tours per day, but in fact, using the data obtained from observation of gravitational and EM fields' distortions, our best scientists stated, that the space-time gate opened with completely different frequency...what is it? experiment is inconsistent with the theory... bad experiment? no, this time - bad theory...
According to the latest intelligence, gathered on Arcon, his magic watch had slightly more cumbersome mechanics...

Imagine, that it is 12:00 now. If you swap minute and hour axis - you will still receive 12:00. But, say it is 03:00. If you swap those axis right away - you'll get a configuration that is never obtained in reality (just try it out...)
Back to Arcon's watch... In fact, the gate opened each time, when there was a possibility to swap the axis in such a way, that the resulting time does make sense!
The question is unchanged:
How many times per day, did the gate open?

Last edited by a moderator: Apr 20, 2004
2. Apr 20, 2004

### gnome

Twelve times each hour from midnight to 11AM and from Noon to 11PM, plus 11 times between 11AM and Noon and 11 times between 11PM and midnight.

Total 12x22 + 11x2 = 286 times per day.

Last edited: Apr 20, 2004
3. Apr 20, 2004

### davilla

Note that any time the hands cross, they can be switched, so there are at least as many solutions to this problem as the last.

Let t be the number of minutes that have passed since the hands are aligned at midnight or noon, in the range [0, 720). The minute hand m = t mod 60, or t = m + 60H for some integer H from 0 to 11. Clearly H corresponds to the hour, or the number of times the minute hand has made a 60-minute revolution. Let h be the minute position of the hour hand. With linear movement, h = t / 12 since h approaches 60 as t approaches 720.

The problem requires that for a given time t there is a hypothetical time q such that the role of h and m in the equations are switched: m = q / 12 and q = h + 60Q for some integer Q between 0 and 11. In fact, Q is the floor of q / 60. The position of m is sufficient to fix a time q, and the position of h determines whether the switch is possible.

Substituting, q = t/12 + 60Q and t = q/12 + 60H, so Q + 12H = 143t / 720. Although Q and H are integers, t is not. Every combination of Q and H gives a different solution for t except that Q = H = 11 is out of range. We should also check that Q is the floor of q / 60 = (12H + 144Q) / 143. Again, this is true for every combination except (11, 11).

So in 12 hours there are 143 times at which the hands can be switched, corresponding to 11 times at which they cross, with t = q and hence H = Q, and 66 = (12 choose 2) symmetrical pairs, t and q of different value, and likewise H and Q. Looked at another way, they are correctly aligned 12 times between crossings, the first and last of which are a symmetrical pair.

Last edited: Apr 20, 2004
4. Apr 20, 2004

### gnome

Or, in plain English:
Within every five-minute period, as the minute hand moves between each pair of numbers, it passes a position where it has traveled a fraction of the distance between those two numbers which is equal to the fraction of a full circle that the hour hand has traveled since 12:00 (except between 11 and 12, because the next corresponding point after 11 occurs at exactly 12:00, which belongs to the next 12-hour period.)
At each of these points, the roles of the two hands can be reversed.

5. Apr 20, 2004

### hemmul

143 - is a correct answer :)
well done again!

However graphical solution is easier:
imagine:
x axis ---> hours
y axis ---> minutes
plotting the possible positions of hour/minute axes, then inverting the coordinate axes and plotting the possible positions again, we recieve the answer as a number of crossings between the two plots. it'd be 143.

6. Apr 20, 2004

### davilla

Wow, this works! It only takes some explanation why switching the hands at these times is legitimate. On the hour, the minute and hour hands point to the start of the full circle and the start of the interval between two numbers, respectively. In one hour, they travel linearly to the end of each interval. This defines a set of all valid positions. The statement you gave is equivalent to saying that the roles of the hands, as I have clarified them, are reversed, just as you claim. The minute hand is evaluated in the smaller interval of the hour, and the hour hand in terms of minutes.

Last edited: Apr 20, 2004
7. Apr 20, 2004

### gnome

hemmul:
Your question was how many times per day...

If a day has 24 hours, the answer is 286, isn't it?

8. Apr 21, 2004

### hemmul

sure, you're right... i just meant that the solution was correct, and... you know i think for physicists it is enough that the order of the result is correct :)

9. Apr 21, 2004

### gnome

Yeah, yeah...

I've had about enough of that bs in my modern physics class. :tongue: