**Ask Marilyn ®** by Marilyn vos Savant is a column in *Parade Magazine*,
published by PARADE, 711 Third Avenue, New York, NY 10017, USA.
According to Parade, Marilyn vos Savant is listed in the
"Guinness Book of World Records Hall of Fame" for "Highest IQ."

In her *Parade Magazine* column of October 1, 2000, a reader poses a puzzle
that Marilyn correctly solves using seven steps.

In her *Parade Magazine* column of November 19, 2000, a different reader
congratulates Marilyn for solving with "pure logic" a problem that is "**not**
solvable in a straightforward mathematical way" because it "appears to have five
unknowns and only four equations."

Rather than point out that the problem **is** in fact solvable in a straightforward
mathematical way using five equations of five unknowns, Marilyn neglects to challenge
the reader's claims, and points out that she's heard from 462 readers who believe her
answer is wrong.

As much as I truly enjoyed the motorboat problem **and** as much
as I also prefer conceptual (what you seem to call "logical") to
strictly mathematical solutions, there is certainly no mystery
about the mathematical solution to this problem. Are you truly
under the impression that it requires solving a set of four
equations for five unknowns and that, somehow, "logic" allows you
to succeed where math fails? Or was your answer simply designed
to lure more incorrect solutions from "mathematicians and other
professionals"?

If the former, please allow me to disabuse you of that silly notion ...

The first time the boats pass we have

dA1 + dB1 = w, (Eq 1)

where dA1 and dB1 are the distances traveled by the two boats and where w is the width of the river.

We can assume (without loss of generality) that boat A is the one that is heading for the shore that is 700 yards away. Thus,

w - dA1 = 700. (Eq 2)

Since you specified that both boats "continue to the opposite shore, then turn around and start moving toward each other again," then, when they pass the second time we have

dA2 + dB2 = 3 w, (Eq 3)

where dA2 and dB2 are the new (longer) distances traveled by the two boats.

Because of the previous assumption it now **will** be boat A that is
300 yards from the shore. So

2 w - dA2 = 300. (Eq 4)

Finally, since the boats travel at constant speeds,

dA2/dA1=dB2/dB1. (Eq 5)

Thus, we have five equations in five unknowns.

Because of the nonlinearity introduced by the constant speed
constraint, there **are** two **mathematical** solutions:

w = 0 dA1 = -700 dB1 = 700 dA2 = -300 dB2 = 300

and

w = 1800 dA1 = 1100 dB1 = 700 dA2 = 3300 dB2 = 2100.

However, we easily rule out the first **mathematical** solution on
**physical** grounds and are left with the second solution. Q.E.D.

It may be interesting to note that, **if** you had not specified
that each boat reaches its opposite shore before the second
passing, Eq 3 could just as easily read

dA2 - dB2 = w, (Eq 3A)

implying that boat A moves faster and catches boat B before B reaches B's opposite shore, or

dB2 - dA2 = w, (Eq 3B)

implying that boat B moves faster and catches boat A before A reaches A's opposite shore.

In the first case the **physical** solution would have involved the
number R1 = 12 + sqrt(102) = 22.0995... and would have been

w = 100 R1 = 2209.95... dA1 = 100 R1 - 700 = 1509.95... dB1 = 700 dA2 = 200 R1 - 300 = 4119.90... dB2 = 100 R1 - 300 = 1909.95...

and in the second case the **physical** solution would have involved
the number R2 = [19 + sqrt(235)]/3 11.4432 ... and would have been

w = 100 R2 = 1144.32... dA1 = 100 R2 - 700 = 444.32... dB1 = 700 dA2 = 200 R2 - 300 = 1988.64... dB2 = 300 R2 - 300 = 3132.97...

http://www.wiskit.com/marilyn/river.html last updated November 25, 2000 by herbw@wiskit.com