I’m a bit flat out, and working on another post, but here’s a quick one. It is a lovely problem told to me by Ed Barbeau, who credits it to Andrei Toom:
Two ladies, A and B, live in separate towns. At the same time one day, each sets out on foot for the town of the other, walking at her own constant rate. They pass at noon. A reaches her destination at 4 pm, while B reaches her destination at 9 pm. At what time did they set out?
UPDATE (29/5)
This seems an easy problem to get tangled, although it’s not too bad if you set up the diagram as above. Then, the ratios of times are the same for each part of the trip (because the ratios of speeds are the same). That leads to the equation
T/9 = 4/T
So, T2 = 36, and thus T = 6 (and also T = –6, if we permit time travel).
I just wanted to point out an alternative, very natural approach that one of my students thought up. (He framed it as a guess-and-check thing, but it is natural to frame it as a proper algebraic solution.) What the student did was consider the fractions of the trips, for each lady, in terms of T. There are various ways to do this, and here is one way.
Consider Lady A on the the first part of her trip. She covers T/(T + 4) of the trip. Similarly, Lady B covers T/(T + 9) of the trip. But together these fractions add up to the whole trip. So,
T/(T + 4) + T/(T + 9) = 1
Then, solve away: multiply both sides by (T + 4)(T + 9), expand and cancel, and we end up with T2 = 36, as before.
Of course, this is a lot messier. (The mess is lessened by adding the second halves of the trips.) Nonetheless, this seems a conceptually simpler approach, even if it is more difficult to think of.
