Solution to Problem #6

The desired times are

	3:16:16 256/719 P.M. and
	8:43:43 463/719 P.M.

at which point the hands are 1/719 of the circumference apart.

Let the clock have circumference 1 and let s(t), m(t), and h(t) represent
the position of the hands-valued on the interval [0,1)-at a time t between 0
= 12:00 noon and 1 = 12:00 midnight (this allows us to write h(t) = t).
Since we desire a solution between 1:00 and 11:00, the possibility that one
hand is close to 0, while another is close to 1 need not be considered.

We want to minimize

	f(t) = max{ |s(t) - m(t)|, |s(t) - h(t)|, |m(t) - h(t)| }

Since this is a piecewise linear function, its minimum lies at a corner,
implying at least one of the three quantities must be zero.

Notice that

	h(t) = t; m(t) = frac(12 t); s(t) = frac(720 t)

where frac(x) = x - floor(x).

Notice also that

	h(t) - m(t) = 0 iff t = k/11;
	h(t) - s(t) = 0 iff t = k/719;
	m(t) - s(t) = 0 iff t = k/708;

for some integer k, so the desired minimum must occur when t is rational
with a denominator that divides one of 11, 719, or 708.

If we can find a solution where the hands are 1/719 of a circumference
apart, then we won't have to worry about the other two cases, as they must
be at least 1/708 or 1/11 apart otherwise.

Let's consider when h(t) = s(t).  In this case, t = k/719 for some k between
0 and 718 and m(t) = frac(12k/719).  Thus, we want
frac(11k/719) to be close to either 0 or 1.  11 and 719 are relatively
prime, so using the Euclidean algorithm, we can find an m and a k such that

	719 m - 11 k = 1.

We note that 719(3) = 2157, and 11(196) = 2156, so

	frac(11(196)/719) = 1/719.

Without breathing hard, we'll find that 719(8) - 11(523) = -1 as well.

The hands have a separation of 1/719 of the circumference at

	t = 196/719 and at t = 523/719

which correspond to times of

	3:16:16 256/719 PM    and
      8:43:43 463/719 PM