A clock has an hour hand of length 3 and a minute hand of length 4. From 1:00 am to 1:00 pm of the same day, find the number of occurrences when the distance between the tips of the two hands is an integer.
In a cycle from 12:00 to the next time the two hands get together, the hands integral distance values go from 1 through 2, 3, 4, 5, 6 and 7, then 6, 5, 4, 3, 2, back to 1 again. We can count only 12 of these 13 as the 1 belongs to two such cycles.
Another way of saying this is that in each 12/11 of an hour, There's one time that the hand tips are 1 unit apart, one time 7 units apart and two times each that they are 2, 3, 4, 5 or 6 units apart.
From 1:00 am to 1:00 pm, the clock goes through 11 such cycles, regardless of how you offset the start and end, as it doesn't matter if you take the end of one cycle and tack it onto the beginning of the next, or vice versa.
That makes the total 12*11 = 132.
To make this rigorous we need to show that at 1:00 such an integral position does not exist; if it did there'd be an extra occurrence as such a situation belonging to two cycles would be present.
At 1:00 the hands are 30° apart. The hand tips are separated by d where
d^2 = 4^2 + 3^2 - 2*4*3*cos(30°) ~= 4.2153903091735 and
d ~= 2.05314157066031, which is not an integer. QED.
Edited on November 9, 2020, 9:57 am
Posted by Charlie
on 2020-11-09 09:55:43