At what time will the three hands of a clock (hour, minute, second) trisect the face? [Hint: Try to prove they never trisect the face.]
[Problem submitted by Roger Wolf, LACC Chairman of Mathematics.]
Solution for Problem 10:
Let h, m and s denote the hour hand minute hand and second hand. If the hands trisect the face, then there are only two possible cases: either m is 20 units ahead of h (the face is divided into 60 units) as in Figure a, or m is 20 units behind of h as in Figure b.
Assume all the hands start at . When the hands trisect the face, h moved x units, m moved 12x units and s moved 60*12x = 720x units.
For Figure a, the amount moved by s plus 20 units will be m turns more than the amount moved by h.
(1) 720x + 20 = x + 60m 719x + 20 = 60m
The amount moved by m subtract 20 units will be n turns more than the amount moved by h.
(2) 12x – 20 = x + 60n 11x – 20 = 60n
Eliminating x from equation (1) and (2), that is, 11*(1) – 719 *(2), we get
11*719x + 11*20 - (719*11x - 719*20) = 11*60m – 719*60n
That is 20*730 =60*(11m – 719n) 730 = 3*(11m – 719n).
3 is a factor of 730. False!
The hands cannot trisect the face in Figure a.
For Figure b, all we have to do is change the signs of 20 in equation (1) and (2). Then we will get the same conclusion.