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OT Analog clock math problems

Posted By: Bob Koca
Date: Tuesday, 4 May 2010, at 6:26 a.m.

In Response To: OT Analog clock math problems (Jason Lee)

For problem 2:

1) First note that the second hand travels 360 degrees per minute, the minute hand travels 6 degrees per minute, and that the hour hand travels 0.5 degrees per minute.

2) Thus note that the difference in the angle between minute hand and hour hand changes by 5.5 degrees per minute and that the difference in the angle between second hand and hour hand changes by 0.5 degrees per hour.

3) Note that at exactly 4:00 and 0 seconds that the hour hand and minute forma 120 degree angle and that the hour and second hand from a 0 degree angle.

4) From that time on the minute hand and hour hand will next form a 120 degree angle after 240/5.5 = 480/11 minutes. (the angle will need to decrease to 0 and then build up again to 120 degrees). After that time there will next be a 120 degree angle formed between hour and minute hand after 120/5.5 = 240/11 minutes. (the angle will need to increas 120 degrees from 120 to 240). From here on that pattern will continue, 480/11 then 240/11 ... giving times where the minute and hour hand form a 120 degree angle.

5) Let's look at every other occurence happening every 720/11 minutes after 4pm. For there to be a chance of the second hand also forming a 120 degree angle with each of the other two hands then it must in particular from a 120 degree angle with the hour hand. For this to have happened the angle must have increased by some multiple of 120 degrees.

6) So after n(720/11) minutes for an integer n, we will have needed a change in angle of x(120).

After n(720/11) minutes the angle change between hour and second hand is .5n(720/11).

7) If you set .5n(720/11) = 120x and solve for n you obtain n = 11x/3

Since n is an integer we need x to be a multiple of 3. But then the angle between hour and minute hand changed by 360 degrees. Therefore it is the same as it was at 4:00, namely 0 degrees.

8) Thus if there is a time that works it will be after n(720/11) + 480/11 minutes.

Setting n(720/11) + 480/11 = 120x and solving for n gives n = 11x/3 - 2. Again we must have that x is a multiple of 3.

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