The ferries over Mississippi

[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.