This a problem that was found on a contest from 1996.
Two swimmers start out from opposite ends (red side and blue side) of a pool. They cross each other for the first time at line PO which is 32 yards from the red side. They reach the opposite ends, turn around and cross again at line RS which is 18 yards from the blue side. Assuming that thye both travel at constant speeds, how many yards is it from the red side to the blue side?
Assuming the swimmers swim at a constant speed, we can conclude that the ratio of the distances traveled is constant. If D is the distance in question, then at the first crossing, the red swimmer has swum 32 yards and the blue swimmer has swum D - 32 yards. At the second crossing, the red swimmer has swum D + 18 yards, and the blue swimmer has swum 2*D - 18 yards. Therefore:
32 / (D-32) = (D+18) / (2*D-18)
A little arithmetic follows:
32 * (2*D-18) = (D-32)*(D+18)
64*D - 576 = D^2 - 14*D - 576
D^2 - 78*D = 0
D = 78 yards
To verify, according to my computations, at the first crossing, the red swimmer has traveled 32 yards and the blue swimmer has traveled 46, a ratio of 16/23. At the second crossing, the red swimmer has traveled 96 yards and the blue swimmer has traveled 138, also a ratio of 16/23. The second set of distances is three times the first set.
The answer seems to be consistent with the conditions of the problem, so I'll guess I got it right. Here's hoping I haven't screwed up the arithmetic somewhere. :-)
Let L = length of pool (the answer you seek)
t1 = time for both swimmers to cross at PO
t2 = time for both swimmers to cross at RS
vr = speed of swimmer starting at red side
vb = speed of swimmer starting at blue side
Write equations for distance = speed * time
Distance travelled by swimmer from red side to crossing at PO:
32 = vr t1
Distance travelled by swimmer from blue side to crossing at PO
L - 32 = vb t1
Distance travelled by swimmer from red side from crossing to crossing at RS
L + 18 = vr t2
Distance travelled by swimmer from blue side to crossing at RS
L + L - 18 = vb t2
The first pair of equations give vb/vr = (L - 32)/32
Second pair gives vb/vr = (2L - 18)/(L + 18)
Very nice observation -- even brilliant. At first I didn't see your point, but after thinking about it, it's a very nice observation. If I could give the points to you, I would. Some solutions are better than others, and in this case your solution is better.
32 / (D-32) = (D+18) / (2*D-18)
A little arithmetic follows:
32 * (2*D-18) = (D-32)*(D+18)
64*D - 576 = D^2 - 14*D - 576
D^2 - 78*D = 0
D = 78 yards
To verify, according to my computations, at the first crossing, the red swimmer has traveled 32 yards and the blue swimmer has traveled 46, a ratio of 16/23. At the second crossing, the red swimmer has traveled 96 yards and the blue swimmer has traveled 138, also a ratio of 16/23. The second set of distances is three times the first set.
The answer seems to be consistent with the conditions of the problem, so I'll guess I got it right. Here's hoping I haven't screwed up the arithmetic somewhere. :-)