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

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.