What is the distance between the tips at the moment when it is increasing most rapidly?

OK, this time I just started with the law of cosines getting d as a function of theta, then took d' and d" and I got the same answer as Charlie's original post.  But I got it without using the solver on the spreadsheet.  Call the angle x, it's too hard to figure out how to write a theta

d=sqrt(25-24cos(x))

d'=12sin(x)/d

setting d" to zero gives:

0=cos(x)sqrt(25-24cos(x)) - 12 sin^2(x) / sqrt(25-24cos(x))

multiply both sides by sqrt(25-24cos(x)) gives:

cos(x)(25-24cos(x)) - 12 sin^2(x)
25 cos(x) - 24 cos^2(x) - 12 sin^2(x)
25 cos(x) - 12 cos^2(x) - [12 sin^2(x) + 12 cos^2(x)]
25 cos(x) - 12 cos^2(x) - 12   or

cos^2(x) - (25/12) cos(x) + 1 = 0
solving the quadratic gives 2 values for cos(x):  3/4 and 4/3 but only 3/4 can be valid since the cosine can't be greater than 1.

so cosine is .75
the angle is arccos(.75) = .7227 radians = 41.41 degrees
distance is 2.646 = sqrt(7)
d' = 3
and there is a right angle between the hour hand and a line connecting the tips of the 2 hands

Better late than never, and that's my final story