Teaser 3025: Please mind the gap
From The Sunday Times, 13th September 2020 [link]
Ann, Beth and Chad start running clockwise around a 400m running track. They run at a constant speed, starting at the same time and from the same point; ignore any extra distance run during overtaking.
Ann is the slowest, running at a whole number speed below 10 m/s, with Beth running exactly 42% faster than Ann, and Chad running the fastest at an exact percentage faster than Ann (but less than twice her speed).
After 4625 seconds, one runner is 85m clockwise around the track from another runner, who is in turn 85m clockwise around the track from the third runner.
They decide to continue running until gaps of 90m separate them, irrespective of which order they are then in.
For how long in total do they run (in seconds)?
[teaser3025]
S2T2 
Jim Randell 5:28 pm on 11 September 2020 Permalink |
For the distances involved they must be very serious runners.
I amused myself by writing a generic function to check the runners are evenly spaced. Although for the puzzle itself it is possible to use a simpler formulation that does not always produce the correct result. But once you’ve got that sorted out the rest of the puzzle is straightforward.
This Python program runs in 55ms.
Run: [ @replit ]
from enigma import (irange, div, tuples, printf) # one lap of the track is 400m L = 400 # time for separation of 85m T = 4625 # check for equal separations def check_sep(ps, v): k = len(ps) - 1 # calculate the remaining gap g = L - k * v if not (k > 0 and g > 0): return False # calculate positions of the runners ps = sorted(p % L for p in ps) # calculate the distances between them ds = list((b - a) % L for (a, b) in tuples(ps, 2, circular=1)) # look for equal spacings return (g in ds and ds.count(v) >= k) # consider speeds for A for A in irange(1, 9): # A's distance after T seconds dA = A * T # B's distance after T seconds dB = div(dA * 142, 100) if dB is None: continue # A and B are separated by 85m or 170m if not (check_sep((dA, dB), 85) or check_sep((dA, dB), 170)): continue # now choose a whole number percentage increase on A for x in irange(143, 199): # C's distance after T seconds dC = div(dA * x, 100) if dC is None: continue # A, B, C are separated by 85m if not check_sep((dA, dB, dC), 85): continue printf("A={A} -> dA={dA} dB={dB}; x={x} dC={dC}") # continue until separation is 90m for t in irange(T + 1, 86400): dA = A * t dB = div(dA * 142, 100) dC = div(dA * x, 100) if dB is None or dC is None: continue if not check_sep((dA, dB, dC), 90): continue # output solution printf("-> t={t}; dA={dA} dB={dB} dC={dC}") breakSolution: They run for 7250 seconds (= 2 hours, 50 seconds).
After which time A will have covered 29 km (about 18 miles), B will have covered 41.2 km (about 25.6 miles), and C (who runs at a speed 161% that of A) will have covered 46.7 km (about 29 miles), which is pretty impressive for just over 2 hours.
For comparison, Mo Farah on his recent world record breaking run of 21,330 m in 1 hour [link], would not have been able to keep up with C (who maintained a faster pace for just over 2 hours), and would be only slightly ahead of B at the 1 hour mark.
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