Why doesn't this work? -- The V.I. Arnold Primary School Problem (Two women started at sunrise...)

Question

I've got a, perhaps silly, question. I understand the various solutions to this problem:

Two women started at sunrise and each walked at a constant velocity. One went from A to B and the other from B to A. They met at noon and, continuing with no stop, arrived respectively at B at 4 p.m. and at A at 9 p.m. At what time was the sunrise that day?

I'm not in need of a solution. I'd like to know why my way of thinking of it doesn't work, though. (For some reason, it is often in mathematics that, what ends up confusing me isn't how to do the problem—but how not to do it.)

Here's my idea:

We know that if the women were both going the same speed—the average speed between them, at 12:00, they would meet up in the middle, right in between A and B. To find the average speed, we consider the situation from the point at which they originally met: It took the first woman 9 hours to finish her trip and the second woman 4 hours. We increment the time it takes them (synonymous with incrementing their velocities) in 30-min. intervals until it takes them the same amount of time. That is:

• The first woman takes 9 hours while the second takes 4.
• The first woman takes 8.5 hours while the second takes 4.5. (This corresponds to speeding up the first woman a little bit, and slowing down the second the same amount, so that their 12:00 meeting point shifts a bit towards the center).
• The first woman takes 8.0 hours while the second takes 5.
• The first woman takes 7.5 hours while the second takes 5.5.
• The first woman takes 7 hours while the second takes 6.
• Both women take 6.5 hours.

So, to get to the center of the path at 12:00 going the average speed between the two would take 6.5 hours. The sunset would have began, then, at $12 - 6.5 = 5.5 = 5:30$. The answer is obviously not 5:30.

Where does this method of thinking go wrong?

Edit: This "solution" doesn't assume that they meet in the middle. It only assumes they meet in the middle if they were both going the average speed between them. Incrementing the time equally (and thus their velocities equally) is supposed to ensure that their meeting time is still 12:00.

Answer 1

They don't meet in the middle, but further away from the starting point of the faster one. And one thing that is often the cause of erorr in such problems is to take the wrong kind of average of velocities: Sometimes arithmetic mean is correct, sometimes harmonic mean, sometimes we better make a sketch. In fact, a sketch (location-time-diagram) is the best starting point for such problems.

Btw. the problem statment is problematic in itself: Since the women start at different locatons, they very likely have different times of sunride (and might be in different time zones, but that would really be nitpicking)

Answer 2

Your claim is that for any (reasonable) difference between $t_1$ and $t_2$, we have that $\frac12(r_1+r_2)$ is a constant. I argue that the claim you're making is quite strong. Even if it were true [and it turns out it isn't], most of the solution would probably be wrapped up in it.

For example, the claim is not true for a fixed distance: from $d=rt$ we have that for any $\alpha>0$: $$\frac12(r_1+r_2)=\frac12\left(\frac{d}{t_1}+\frac{d}{t_2}\right) \neq \frac12\left(\frac{d}{(t_1+\alpha)}+\frac{d}{(t_2-\alpha)}\right).$$

However, that fails to solve the problem posed, because, as you note, we do not assume that for all $\alpha$, the two women meet in the middle. Therefore to plug in $t_1=t_2=6.5$ and $\alpha=2.5$, we must assume that $d$ refers to the distance between where they meet and their final destinations, but that is not the same for both women.

Therefore, let $d$ be the distance from $A$ to $B$ and suppose they meet at a proportion $\beta$ of the way along the path. Then we are trying to compare $$\frac12\left(\frac{d}{t}+\frac{d}{t}\right) ~~\text{and}~~ \frac12\left(\frac{\beta d}{(t+\alpha)}+\frac{(1-\beta)d}{(t-\alpha)}\right).$$

Now, it's not clear that $\beta$ as a function of $\alpha$ won't magically work to cancel things out, but it turns out that it doesn't. There is probably some good reason for this but I can't figure it out.