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So I was facing a Riddle from Dudeney's Canterbury Puzzles book (number $99$), which states :

A country baker sent off his boy with a message to the butcher in the next village, and at the same time the butcher sent his boy to the baker. One ran faster than the other, and they were seen to pass at a spot $720$ yards from the baker's shop. Each stopped $10$ minutes at his destination and then started on the return journey, when it was found that they passed each other at a spot $400$ yards from the butcher's. How far apart are the two tradesmen's shops? Of course each boy went at a uniform pace throughout.

I was trying to write down the equations of motion, but I realised that I cannot.

Calling $A$ and $B$ the distance between baker and butcher, I named a point $P$ where they meet the first time.

Because the first (let's choose it) runs faster, it will surely, over a time $t$, walk more yards than the other. And also it will arrive first at the destination.

Yet the other may still be walking to reach his destination. But whilst he is still walking, the other is resting and those $10$ minutes decrease. At the end of the first person's rest, the other may or mayn't still walking, or he may be resting or why not he may already have to arrive.

How do I set up the equations for a problem like this?

For the records, I report the "apparently weird" solution (weird because the method is weird, and there is no explanation).

Solution

All that is necessary is to add the two distances at which they meet to twice their difference. Thus $720 + 400 + 640 = 1760$ yards, or one mile, which is the distance required. Or, put another way, three times the first distance less the second distance will always give the answer, only the first distance should be more than two-thirds of the second.

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    $\begingroup$ I don't see such a simple way to get the answer, though I get the same result. My method: Say they first meat at time $1$ (that just sets a time unit). Then $v_1=720,v_2=d-720$ ($d=$ the answer). Ignoring the waiting period (as it has no bearing on anything) and equating the times to the next meeting we get $\frac {720+d-400}{d-720}=\frac {d-720+400}{720}\implies d=1760$ $\endgroup$ – lulu Jan 9 '18 at 20:31
  • $\begingroup$ Note: the problem clearly states that both have arrived,tarried, and started home prior to the second meeting, so I am taking that as granted. $\endgroup$ – lulu Jan 9 '18 at 20:32
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Call $d$ the total distance, $d_1$ the "first" distance and $d_2$ the "second" distance. Now, call $v_1$ the speed of one of the boys and $v_2$ the speed of the other one (the first boy is running the distance $d_1$ until they meet for the first time). We have:

$$\frac{d_1}{v_1}=\frac{d-d_1}{v_2}$$

(the time until they meet the first time), and

$$\frac{d+d_2}{v_1}+10\text{min}=\frac{d+(d-d_2)}{v_2}+10\text{min}$$

(the time until they meet the second time).

Cancelling $10\text{min}$ and shuffling the factors gives us:

$$\frac{d_1}{d-d_1}=\frac{v_1}{v_2}=\frac{d+d_2}{2d-d_2}$$

thus:

$$d_1(2d-d_2)=(d-d_1)(d+d_2)$$ $$2dd_1-d_1d_2=d^2-dd_1+dd_2-d_1d_2$$ $$0=d^2-3dd_1+dd_2$$

and, assuming $d\ne 0$: $$d=3d_1-d_2$$

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