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Here's an old challenge from an old book (I think it goes like this, if I remember well):

There is a place on Mississippi river between Missouri and Kentucky where are two ferry boats: the vintage Old Town and the modern president. One time, exactly at the same time, the Old Town left the Missouri shore and the president left the Kentucky shore.

The ferries over Mississippi

Each ferry traveled at a fixed velocity and, after a while, they intersected each other 9000 feet from Missouri. When they arrived to the other side, they didn't waste any time and immediately started the journey back to intersect each other 3000 feet from Kentucky.

QUESTION #1:
What is the width of the river on the site where this story takes place?

QUESTION #2:
How far away from Kentucky will they intersect for the third time?

EDIT: For those who voted to close this question, this is NOT a mathematics problem. It's a mathematics puzzle. At least it's from an old book of puzzles (which title I honesty can't remember). You can't solve it by the use of conventional formulas like velocity or others. Can you at least leave a comment?

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    $\begingroup$ I'm pretty sure this puzzle was in a Martin Gardner book, based on his SciAm columns. The numbers 3000 and 9000 seems a bit off, since the puzzle is about ferries crossing the river, not going up and down the river. $\endgroup$ – Jaap Scherphuis Jul 20 at 15:22
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    $\begingroup$ What makes this a nice puzzle is that with the right insight, it becomes really easy. While you could do it by solving a set of simultaneous equations, there is no need to do that. $\endgroup$ – Jaap Scherphuis Jul 20 at 15:56
  • $\begingroup$ @JaapScherphuis, thank you for your insight. I think you are right. Although the book I remember was not a Martin Gardner book, I think the problem had that reference. And because it was a long time ago, I can't recall if the numbers 3000 and 9000 are exactly right. Maybe it was 300 and 900... If you want to edit the values on the question (as well on your answer), please do so! $\endgroup$ – Pspl Jul 20 at 22:05
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Here is a simple argument to deduce the solution to the first question without much arithmetic.

When the boats first meet, the total distance they have travelled is the width of the river.
When boats meet again, their total distance is three times the width of the river (each has done a full crossing, and they meet on their return trips which together span the river).
The boat leaving Missouri has travelled 9000 feet at the first meeting. By the time of the second meeting it will have travelled three times as far, or 27000 feet. We know that it has crossed the river once and are given that it travelled a further 3000 feet from that opposite side, so the river must be 24000 feet wide.

The second question is not quite as nice, but now we know the width of the river and the relative speed of the boats, so it can all be calculated fairly easily.

If we wait till the boats once again meet each other while going in opposite directions, their total distance will be 5 river widths. The first boat will therefore have travelled $5\cdot9000=45000=24000+21000$ feet. This means it is still on its first return trip, just $3000$ feet from its starting point in Missouri, $21000$ from Kentucky.
Note that this means that the other boat has done just over three river widths, on its second return trip. It must have actually overtaken the first boat at some point while going in the same direction towards Missouri. While the first boat does $9000$ feet, the other does $24000-9000=15000$ feet, so we can calculate exactly when everything happens. It turns out to occur when the first boat has done $4\cdot9000=36000=24000+12000$ feet, and the other has done $4\cdot15000=60000=3\cdot24000+12000$ feet. So the first boat gets overtaken exactly in the middle of the river.

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[Revised to cover both parts of the puzzle, beginning with the original post.]

While the puzzle has already been completely solved by Jaap Scherphuis a geometric approach works very nicely for the puzzle’s first part that asks for the width of the river. As time flows downward (as well as downstream if the ferries are carried by a current, thanks to Pspl’s observation, the same diagrams apply) on these graphs the trajectory of the slower ferry looks steeper than that of the faster ferry.

Begin by representing the first and second meetings of the ferries. Notice that the first meeting forms a triangle.

Now unfold the graph of the second meeting by interpreting each ferry’s turnaround at a bank as reflection off of a mirror. Note the large triangle formed by this unfolded second meeting.

Flipping that second-meeting triangle horizontally reveals it to be congruent to the first-meeting triangle and 3 times its size. The calculation of the river’s width may now be completed by breaking 3 copies of the first triangle and rearranging the pieces.

[The rest was added a day later.]

To find the third meeting point begins with unfolding a diagram that assumes that the faster ferry will overtake the slower ferry after bouncing off the Kentucky shore.

This is now equivalent to a race where the faster ferry catches the slower ferry, which begins with a one river-width head start. The meeting point may be found by forming a parallelogram whose base is the by-now-known width of the river, sides are the path of the slower ferry and long diagonal is the faster ferry’s path.

There’s the first meeting’s triangle again, along with its left-to-right reflection, helping to form and measure a pair of right triangles geometrically similar to those that define the parallelogram’s sides and long diagonal.

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    $\begingroup$ Have you just prove that the stream of the river doesn't affect the result?! Thumbs up! $\endgroup$ – Pspl Jul 21 at 10:43
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    $\begingroup$ Excellent observation, @Pspl ! It has been added to the post. $\endgroup$ – humn Jul 21 at 19:51

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