mows
Posts:
4
Registered:
12/13/04


Re: clock puzzle...
Posted:
Jul 31, 2003 2:41 PM


"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

