Saturday, 10 January 2009

SOLUTION to Al's Clock Problem


Here's my solution to Al's Problem about the Concert Clock. My strategy was to find the angle of the hour hand between six and seven, and set it equal to the angle of the minute hand between nine and ten. I found that the concert starting at 6:49 1/11 and ending at 9:32 8/11.

Here is how I came up with that... The number of degrees rotation from straight up (12) on the clock for the minute hand is 6T where T is in minutes past the hour, for any hour. The hour hands move T/2 degrees per minute so after nine oclock the angle of the hour hand is at 270 + T/2 degrees for any time T, and between six and seven it is at 180 + T/2. My next step was to express these hour angles in terms of the minute angles...
Using the equation 180+T/2 =H (angle measure of hour hand) and 6T=M (angle measure of the Minute hand) we can solve together to eliminate T and get H=180 + M/12... This is the hour hand angle between six and seven... If we set this equal to the minute hand angle between nine and ten, we will find the angle when they are the same.
Setting 180 + M/12 = 6M and solving I get 196 4/11 degrees.

Now we can use the M = 6T and set it equal to the 196 4/11 and solve for T, The time turns out to be 32 8/11 minutes after the hour, so the ending time is 9:32 8/11.

To find the beginning time, we must set the hour hand angle for nine oclock agains the minute hand for six oclock, to get 270 + M/12 = M. This gives an angle of 294 6/11 degrees. Using the minute hand formula, M = 6T we see that the number of minutes after six oclock is 49 1/11, giving us 6:49 1/11 for the beginning time.

Hope I made that clear

I'm having fun with this, and hope you all are, so maybe I'll post another in a few days... Thanks again to Al for the nice problem.

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