Two Ferry Boats

Martin Gardner, in his book Hexaflexagons, Probability Paradoxes, and the Tower of Hanoi presents the following problem on page 123.

Two ferry boats start at the same instant from opposite sides of a river, travelling across the water on routes at right angles to the shores. They pass at a point 720 yards from the nearest shore. Both boats remain in their slips for 10 minutes before starting back. On the return they meet 400 yards from the other shore.

How wide is the river?

I have left the units in yards for two reasons: firstly, it is the unit that is quoted in the book, and secondly, it adds another interesting element to the final solution.

I puzzled over this for a few days. I had got about half way through the calculation ( I now realise from hindsight ) but couldn't see a way forward. It even kept me awake at night. A few nights ago, at 1:00 am, I hit on the strategy, got up and tried it out on paper; when I checked Gardner's answer I found that we agreed.

My solution is algebraic, and Gardner gives a very brief description of it, but it is also possible to solve this without maths, as Gardiner demonstrates in a long written paragraph.

Try to find the answer for yourself before you "Click to read the rest of the post" for my solution.


Define variables

Two time periods can be defined
T₁ = interval between the two times that the boats pass
T₂ = interval between when both boats "start at the same instant from opposite sides" and the second time that they pass

Five distances can be defined
Distances relating to T₁
D₁ = distance fast boat travels between the two passes
D₂ = distance the slow boat travels between the two passes
Distances relating to T₂
D₃ = distance the fast boat travels from start to second pass
D₄ = distance the slow boat travels from start to second pass

W = the width of the river

The two boat speeds
S₁ = the speed of the fast boat
S₂ = the speed of the slow boat

The diagram below presents the problem graphically.



Using the diagram above we can assign values to the distance variables.

D₁ = 720 + W - 400 = W + 320
D₂ = W - 720 + 400 = W - 320
D₃ = 2W - 400
D₄ = W + 400

The basic strategy for solving this problem can be outlined as follows:
As two time lengths can be found and distances the boats travel in these time frames can be calculated, the ratio of the boat speeds can be defined in two ways, using the distances.


Equate the two ratio equations and solve for width of river (W).

Note: as both boats spend an equal amount of time in the slips (10 minutes) that can be ignored as it doesn't effect the result of the calculation (it is a red herring).

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Calculate the ratio of the boat speeds using distance information for T₁
Note: time = distance / speed

T₁ = D₁ / S₁ = (W + 320) / S₁
T₁ = D₂ / S₂ = (W - 320) / S₂

As both equations = T₁ they can be equated

(W + 320) / S₁ = (W - 320) / S₂


Cross multiply to express in terms of the ratio of the speeds

S₁ / S₂ = (W + 320) / (W - 320)    Equation 1

Calculate the ratio of the boat speeds using distance information for T₂

T₂ = D₃ / S₁ = (2W - 400) / S₁
T₂ = D₄ / S₂ = (W + 400) / S₂

As both equations = T₂ they can be equated

(2W - 400) / S₁ = (W + 400) / S₂

Cross multiply to express in terms of the ratio of the speeds

S₁ / S₂ = (2W - 400) / (W + 400)     Equation 2

We have now defined the ratio of the boat speeds in two ways: Equation 1 and Equation 2

All that remains is to equate Equation 1 and Equation 2 and solve for W (width of the river).

(W + 320) / (W - 320) = (2W - 400) / (W + 400)

Cross multiply to remove fractions.

(2W - 400)(W - 320) = (W + 320)(W + 400)

-> 2W² - 640W - 400W + 128000 = W² + 400W + 320W + 128000
-> 2W² - 1040W + 128000 = W² + 720W + 128000
-> 2W² - W² - 1040W - 720W + 128000 - 128000 = 0
-> W² - 1760W = 0

Extract W

-> W(W - 1760) = 0

-> W = 0 which can be ignored as it is nonsensical in terms of the practical problem (as the function is a parabola it should have two solutions, one of which is at the origin)

or

W - 1760 = 0

W = 1760

Therefore the width of the river is 1760 yards

Can you see the second reason why I kept the problem in yards rather than change the units into the modern value of metres?

Any comments?