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Topic: clock puzzle...
Replies: 17   Last Post: Aug 2, 2003 7:00 PM

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mows

Posts: 4
Registered: 12/13/04
Re: clock puzzle...
Posted: Jul 31, 2003 2:41 PM
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"Michael Hartley" <policymodel@hotmail.com> wrote in message
news://cfcb68e.0307302309.7748ed5e@posting.google.com...
> Dear All,
>
> Here's a tough puzzle:
>
> Suppose a clock has 3 hands of equal length (hour, minute, second).
> Consider the triangle connecting the ends of the hands.
>
> At what time is the area of the triangle the largest?
>
> Enjoy!


For a 12 hr clock face.

Given 3 points (x1, y1), (x2, y2), (x3, y3), the signed area of a triangle
is
[(x1 y2 - x2 y1) + (x2 y3 - x3 y2) + (x3 y1 - x1 y3)] / 2
With s as seconds since 12 o'clock, the coodinates of the end of the hands
are
(Sin [Pi s / 30], Cos[Pi s / 30]),
(Sin [Pi s /1800], Cos[Pi s /1800]),
(Sin[ Pi s / 21600], Cos[Pi s / 21600])
for second, minute and hour hands respectively
The signed area of the triangle is therefore
(Sin[11 Pi s / 21600] + Sin[59 Pi s / 1800] + Sin[719 Pi s / 21600] ) / 2
differentiating and set to 0
(11 Cos[11 Pi s / 21600] + 708 Cos[59 Pi s / 1800] + 719 Cos[719 Pi s /
21600] ) / 2 = 0
maximum area is at
20949.12338408509465052277227452846 sec after 12 and
22250.87661591490534947722772547154 sec after 12.


These are the times
05 hr 49 min 09.12 sec and
06 hr 10 min 50.88 sec


Note, the times add up to 12 hr.
The angles (degrees) between hands are

120.3363102
119.8357237
119.8279661

24 hr clock face
The signed area of the triangle is
(Sin[23 Pi s / 43200] + Sin[59 Pi s / 1800] + Sin[1439 Pi s / 43200] ) / 2
and the max area is at
20033.905599933416723 sec after 00 hr and
66366.094400066583277 sec after 00 hr.

These are the times
05 hr 33 min 53.91 sec
18 hr 26 min 06.09 sec

Note, the times add up to 24 hr.
These are more "equilateral" than the 12 hr face.
the angles (degrees) between hands being

120.043039607
119.915953327
120.041007066

It seems that there is no true "equilateral" time




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