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Re: Tortured Algebra by an amateur



>>> Allen.Flanigan@USPTO.GOV 02/27 2:09 PM >>>
...
Is there a more elegant solution using algebra?
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Not exactly elegant, but kind of neat. I remember it because one of my older stepsons brought this problem home as a "brain-teaser" in his junior math class.

You get it by expressing the distance traveled by the hands aournd the clock as distance around a circle as a function of time (naturally enough.)

'Spose H is the hour hand M is the minute hand.

The hour hand goes around the clock once every 12 hours. So, its position around the circle
after t hours is

H(t) = ( 1/12 turn/hr x 360 deg/turn) x ( t hours),

or H(t) = 30t

The minute hand goes around the clock once every hour, so its position around the circle after t hrs is

M(t) = (1 turn/hr x 360 deg/turn) x (t hours)

or M(t) = 360t

By setting the distance formulas equal to each other, we can find the time when when the hands overlap.

H(t) = M(t)
30t = 360t
0 = 330t
t = 0

(Simplify things by requiring that time(0) equals 12 o'clock.)

Find the other solutions by realizing that for every hour the minute hand travels a multiple of 360 degrees farther than the hour hand.

M(t) = H(t) + 360N
360(t) = 30(t) + 360N
330t = 360N

t = (360 / 330) x N, which is
t = (12 / 11) x N

You can just plug in the values to find the times where the hands overlap.

I think I got something slightly wrong somewhere, but this is what I rememember. Oh, and I won't say whether it was me or my stepson who solved it, heheheh.

Art

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