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.
This is the problem from Vladimir Arnold’s little book”Problems for Children From 5 to 15″, a collection of 77 mathematical problems designed to challenge and stimulate young minds. They say that he (Arnold) solved it when he was 12.
Ah. Thanks, Dr. M. Yes, I see it is problem 5, here or here.
The book is incredible and I highly recommend it!
Here is another incredibly mind boggling problem from the same book: Two volumes of Pushkin, the first and the second, are side-by-side on
a bookshelf. The pages of each volume are 2 cm thick, and the cover – front
and back each – is 2 mm. A bookworm has gnawed through (perpendicular to
The pages) from the first page of volume 1 to the last page of volume 2. How
long is the bookworm’s track?
Reminds me of a poem by Ernst Jandl: “Lechts und Rinks kann man nicht velwechsern – werch ein Illtum”. It translates as “You cannot confuse left and right – what a fallacy”, but the German version lives from switching every l and r in the text.
Also, there’s a high school graduation exam from 2016 that went like this:
At 9:00 Uhr a car starts driving from A to B. After two thirds of the route he passes a bicycle going in the
direction of A. When the car arrives at B, a bus starts driving from B to A. After three fifths of the route
he passes the bicycle. When will the bicycle arrive in A when it is known that the bus arrived at 11:00 h?
(All velocities are assumed to be constant).
Thanks, Franz. It’s an interesting problem. Perhaps I missed a trick, but is there sufficient information to solve the problem?
I first did the special cases when the car or the bus has infinite velocity.
Thanks, but I’m not sure I understand. Are you saying the problem is solvable with such an extra assumption but the answer to the specific question is the same?
Cute.
A pdf is here:
Click to access taskbook_arnold_en_0.pdf
Andre Toom, who worked in Brazil, was for many years a participant in a mathematics education discussion group that I frequent. He gave this problem as an example of the sort of thing commonly given to Russian pupils at the middle school level. His response to the question of how it was possible to regularly assign such problems was essentially that if you keep doing it, the kids get the idea.
I have tried this with a class at this level. If they see it as an algebra problem and try to set up an equation, they are usually dead in the water. It is one of those problems where, if you perceive it wisely and grasp the underlying principle, falls out. One tactic if the kids are stuck is to ask for a guess at the possible starting time; checking this seems to get them into the right frame of mind.
i think that there is a lot of value, even from a DEI point of view, of using a collection of traditional problems from various culture. The complete collection from Alcuin of York is available on line and the ancient Indians were pretty good at making up such problems.
Here’s even another source, 1998 Michigan Autumn Math Challenge:
“Two towns, A and B, are connected by a road. At sunrise, Pat begins biking from A to B
along this road, while simultaneously Dana begins biking from B to A. Each person bikes at
a constant speed, and they cross paths at noon. Pat reaches B at 5 p.m. while Dana reaches
A at 11:15 p.m. When was sunrise?”
I have the pdf since 2006; I’ve tried to find it now and I only could via web.archive:
Past Exams
I have a good collection of sets of problems to which I often refer. If you do enough of these problems, you start to see the same ones over and over again.
Yes. Of course any sane collector will seek to steal from other good collections.
Thanks Marty for sharing this and thanks everyone else for the discussion and pointers to other problems. I am curious about the original problem and (without giving away the answer, please) I am wondering about the different methods of solution.
So far I have tried and found a couple of geometric approaches and a couple of algebraic ones. All lead (some quicker than others) to a quadratic equation (albeit a trivial one in most cases) but I am curious if anyone has found a numerical/algebraic or geometric solution that does not involve the square of the unknown..?
Perhaps it does not exist and in a way this makes sense given the nature of the unknowns as distance and time and the subsequent need to cross-multiply at some point, but I have been surprised before with creative solutions, so… anyone have one?
Thanks, RF. I think the problem is intrinsically quadratic, which is is surprising, and why the problem is surprisingly tricky.
Feel free to give your solution. I’ve added a second diagram, which includes some hinty notation. Apart from a solution that is (with hindsight) natural and easy, I had a student come up with a very nice alternative approach; it ends up being longer, but it’s a cool idea.
OK. Here is my algebraic solution.
Name the women WA and WB (depending on where they depart from). WA walks in 4 hours what WB walks in T hours. WB walks in 9 hours what WA walks in T hours.
So in
hours, WA walks as far as WB does in 
hours.
To walk this distance, WA needs
hours.
This gives
and since we assume 
, 
which means they started walking at 6am.
Thanks, RF. I’d express it maybe a little simpler as
T : 9 = 4 : T
That is the ratio of times are equal, since the ratio of speeds is the same.
I had a student come up with a very different approach, which seemed nicer, albeit leading to messier algebra.
Problems from ancient Greece.
2018-MAV-Mills-Ratcliffe
I’ve updated the post with a couple methods of solving it.