Car A starts at time t = 0 and, traveling at a constant speed, completes 1 lap every hour. Car B starts at time t = α > 0 and also completes 1 lap every hour, traveling at a constant speed. Let a(t) be the number of laps completed by A at time t, so that a(t) = [t]. Similarly, let b(t) be the number of laps completed by B at time t. Let S = {t ≥ α : a(t) = 2 b(t) }. Show that S is made up of intervals of total length 1.
Solution
Answer: S = [ [α]+α, 2[α]+1 ) ∪ [ 2α]+2, [α]+α+2 ).
At all times in the interval [ [α]+α, 2[α]+1 ) car A has completed 2[α] laps and car B has completed [α] laps. At all times in the interval [ 2α]+2, [α]+α+2 ) car A has completed 2[α]+2 laps and car B has completed [α]+1 laps. The first interval has length 1-{α} and the second interval has length {α}, where {α} denotes the fractional part of α.
During the interval [ 2[α]-1, 2[α] ) car A has completed 2[α]-1 laps and car B either [α]-2 or [α]-1 laps, so car A has completed more than twice as many laps as car B. Similarly, during the interval [ 2[α]-n, 2[α]-n+1) car A has completed 2[α]-n laps and car B [α]-n-1 or [α]-n laps, so at all times before 2[α] car A has completed more than twice as many laps as car B and these times do not form part of S.
During [ 2α]+1, [α]+α+1 ) car A has completed 2[α]+1 laps and car B [α] laps, which is less than half the number. During [ [α]+α+1, 2[α]+2 ) car A has completed 2[α]+1 laps and car B [α]+1 laps, which is more than half the number. So no points in the gap between the two intervals of S belong in S.
Finally, during [ [α]+α+2, 2[α]+3 ) car A has completed 2[α]+2 laps and car B [α]+2 laps, whilst during [ 2[α]+n, 2[α]+n+1 ) car A has completed 2[α]+n laps and car B either [α]+n-1 or [α]+n laps. So at all times after [α]+α+2 car A has completed less than twice the number of laps completed by car B and none of these times belong in S.
© John Scholes
jscholes@kalva.demon.co.uk
7 Feb 2001