# Analysis - Clock Hands Meet Puzzle & Generalization to N hands

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Now coming back to point 1, the question was, does the shape matter. Here's what this question means. Imagine a circle with 60 equidistant ticks. Instead of hands of a clock, imagine 2 points moving along the perimeter at different, but uniform speeds. The points start off at 12'o clock just like the hands of the clock. These points would meet at the same position as the clock's hands, dividing the the perimeter into equal parts.

When we asked "does shape matter", we meant, what if instead of a regular circle, the points moved along an arbitrary closed curve. For eg. as the figure depcits.
Question is, will the points still meet so as to cut equal arcs, or equal angles? The answer per my imagination is that they will meet so as to cut equal arc lengths.

TODO: Proof.
As a weak proof, stretch and even out the perimeter of the curve out so that a circle is formed. This case is already (proved) above. Hence points moving along arbitrary closed curve exhibit the same behavior as moving along a circle.

Now, since the shape of the track on which the points are moving does not matter, imagine 2 cars racing side by side on a straight track, at uniform but different speeds. The track has length, say L. The track is a magical track in the sense that as soon as the cars reach the end of the track, the cars disappear from there and instantly appear at the beginning of the track. There on, cars continue racing as if nothing happened.

This situation is similar to a circular track. In this case too, the cars will meet or cross each other at points that divide the track L into equal segments. (The situation is different is the cars travel back from the end to the beginning of the track, as they would do in real life. This is a another interesting situaton to explore, may be later.)

Another conclusion : The shape of the track or the curve does not matter. The moving objects/points meet so as to cut equal arc/segment lengths.

Now lets ask ourselves a more difficult question. How many times and at what times do the hour, min and second hands meet in a 12 hr period.

This may seem complex, but it's not and for the sake of brevity, I leave it to readers to work out a few examples. However, here's the general idea.

When all 3 hands meet, the first condition is that the minute and hour hand should meet. This we have already worked out. Hence the 3 hands MUST meet at one or a few of these points where minute & hr. hands meet. To solve this, we now find out points where minute and second hands meet. Some of these points will coincide with the points where hr. and minute hands meet. These are the points where all 3 of them meet. This situation can also be solved using LCM & GCD. Here's how.

Let's take an example, before we actually go to the conventional clock.
Considering a clock with 60 ticks along its perimeter.
Assume Speed (minute hand) = 7 times speed (hr hand)
Thus these hands meet 6 times and the length of each arc = 60/6 = 10 ticks.

(The hands meet at ticks 10, 20, 30, 40, 50, 60) ---- I

Further assume Speed (second hand) = 10 times speed of (minute hand)
Thus these hands meet 9 times and the length of each arc = 60/9 = 20/3 ticks.

(The hands meet at ticks
20/3, 40/3, 20, 80/3, 100/3, 40, 140/3, 160/3, 60) ----- II

Now we find LCM of 20/3 and 10.

LCM ( fractions ) = LCM (Numerator) / GCD (Denominators)
Hence LCM (20/3 and 10) = LCM ( 20, 10 ) / GCD (1, 3)
= 20/1
= 20

In the figure, The GREEN dots indicate the positions where hour and minute hands meet. The ORANGE dots indicate the positions where minute and second hands meet. Overlap of both the dots indicate the positions where all 3 hands meet.

Thus, all the 3 hands meet at 20 th tick and then 40 th tick and then 60th tick.
This can be confirmed by looking at ticks that are common between I and II.

Observation : The 3 hands also meet so as to cut equal arc lengths.
For the conventional clock,
speed(min hand) / speed (hr hand) = 12
speed(second hand) / speed (min hand) = 60

Hr & Min hand meet every 60/11 ticks.
Min and Second hand meet every 60/59 ticks.
LCM( 60/11, 60/59 ) = LCM (60, 60) / GCD (11, 59)
= 60 / 1
= 60

Thus, the hr, min and second hand meet only once in 12 hr period, i.e at 12'o clock.

Generalizing, some more conclusions:

1. Given a system of n points moving at different but uniform speeds around the perimeter of a closed curve, the points meet at positions so as to cut arcs of equal length in the time that the slowest point goes around the perimeter once.
2. Let the speeds of these n points be S1, S2, S3, S4....Sn, written in decreasing order of magnitude. Each arc length can be calculated using the formula LCM( S1/S2, S2/S3, S3/S4, ... , Sn-1/Sn)
3. If the speed is written in terms of ticks, then the number of times the hands meet can be calculated using 60/LCM( S1/S2, S2/S3, S3/S4, ... , Sn-1,Sn) There may be a better way to find out the 'number of times', but requries more investigation.

Now, we mentioned earlier that there MAY be a relation of these moving 'hands' to Fourier transforms. Here's why. Fourier transform breaks down a given signal into cycles of cycles. Hence if we use the speeds of these hands as input to the transform, it will give rise to a graph. It is possible that the graph will be able to tell us the 'number of times' and the 'at what times'.

Further exploration:

1. This is an interesting problem and several variations of this can be studied (like, going back to car example, what happens when the car instead of disappearing at the end of the track, makes a U turn and goes back from end to the beginning of the track and so forth.)
2. It may be possible to solve the problem by introducing some Physics into the mix (i.e. by considering relative angular speeds, etc.)
3. Some results explained here require formal proofs.
4. Behavior of the system when the speeds are non-integral (fractional, irrational) multiples (e.g minute hand moves sqrt(2) times the hr. hand) can be explored further.
Feel free to share any suggestions/ideas at hereiam(at)rohitn.com.

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