Analysis - Clock Hands Meet Puzzle & Generalization to N hands
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"How many times do clock's hour and minute hands meet in a 12 hr. period? What is the time at the point where they meet?"
This is a problem everyone has heard and tried to solve at some point of time. Few days ago I overheard some teenagers having a conversation about this and the problem revolved in my head again. Though 'how many times' is a simple question to answer, the exact time when they meet is a little tricky. Looking at it as if the minute hand is chasing the hour hand, we run into sum of infinite series.Sometimes, the 2nd part of the question requires you to know the answer to the first part, which kind of feels like "cheating" to get the answer to the second one. :) I decided to give it some more thought and this is what I came across. Here, we also extend our solution to figure out a general case for a clock with N hands.
Before we start, assume that the clock's perimeter is divided into 60 ticks (which are minutes). We can also assume 360 ticks, one for each degree, but it does not matter. We will stick to 60 ticks. For all cases, we will assume that the initial position of the hands is at 12'0 clock before the hands start moving. We will stick to hands or points moving along a 2D plane.
In the conventional clock, the speed of the minute hand is 12 times the speed of hour hand. The hands meet 11 times in 12 hr period. 12 and 11, any relation with speed there? We'll see. Let's try to generalize this problem by assigning different speeds to the minute hand compared to the hour hand.
Hence, in general, if the speed of the minute hand is n times the speed of hour hand, then, we can conclude
Also, an important observation that we made is that the hands meet so as to divide the perimeter into equal parts (the central angle is also equally divided, but we will see later that it is the perimeter that matters, not the angles :) )
So this settles the the matter of the conventional clock. We no longer have to run into infinite series again. The answer is simple.
How many = (12-1) times.
All we need to do to answer the question is to convert these "ticks" into actual time. When the speed of the minute hand is greater than the hour hand, as a general rule, the minute hand has to make one complete rotation every time it has to meet the hr hand. So after starting at 12'o clock, the minute hand will make one complete rotation and then meet the hr hand at 1 hr + 60/11 th tick. Then again, minute hand has to make another rotation and meet the hour hand at 2 hr + 120/11 th tick and so on.
So the hands meet at 1hr + 60/11 mins, 2 hr + 120/11 mins,...11 + 11*60/11 mins (which is nothing but 12'o clock again)
Well, we are not done yet. There's something more we need to explore, however, let's jot down a few points that we need to investigate later.
Now coming to some conclusions: