Problem Setup:
A man is on a pier, holding a rope connected to a boat on the sea.
The man moves backwards (thus pulling the rope by the same amount) one
meter.
How closer does the boat move?

Further clarification:
There are no lengths or heights given.

I'm thinking that as the hypotenuse is decreased, only the bottom leg
decreases with it, as the height is actually the height of the pier.

Hello there:
This is an interesting problem.
Suppose that:
a = height of the pier.
b = distance of boat from pier.
c = length of rope.

Thus, by the Pythagorean theorem, we have:
a² + b² = c²
when the hypotenuse is shortened by one m, we have:
a² + (b - x)² = (c - 1)²
where x represents the distance that the boat had moved.
Now our task is to solve for x:
a² + b² - 2bx + x = c² - 2c + 1
x² - 2bx + a² + b² - c² + 2c - 1 = 0
x² - 2bx + 2c - 1 = 0  (a² + b² - c² = 0)
By using the quadratic formula, we can solve for x:
x = (2b ± squrt(4b² - 4(1)(2c - 1))/2(1)
x = (2b ± squrt(4b² - 8c + 4))/2
x = (2b ± 2 * squrt(b² - 2c + 1))/2
x = b ± squrt(b² - 2c + 1)
Considering the positive part:
The new distance of the boat from the pier is:
b - x = b - (b + squrt(b² - 2c + 1))
      = - squrt(b² - 2c + 1), which doesn't make sense, so we ignore
it.
Considering the negative part:
b - x = b - (b - squrt(b² - 2c + 1))
      = squrt(b² - 2c + 1)

So, to answer your question, when the hypotenuse is shortened by 1 m,
the boat moves b - squrt(b² - 2c + 1) meters closer to the pier,
making its new distance to be squrt(b² - 2c + 1).

Note: squrt denotes the square root symbol.
I hope this answered your question, if you need clarification, please
let me know and I'll be happy to answer.
secret901-ga

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Request for Answer Clarification by zarkon-ga on 05 Sep 2002 19:31 PDT

It seems to have no errors so far.. however, I seem to have forgotten
my basic algebra.

The line:
x = (2b ± squrt(4b² - 8c + 4))/2 
x = (2b ± 2 * squrt(b² - 2c + 1))/2 

What is the rule that allows the factoring out of a 2 and resulting in
actually factoring out a 4?

---

Clarification of Answer by secret901-ga on 05 Sep 2002 20:20 PDT

zarkon:
x = (2b ± squrt(4b² - 8c + 4))/2
x = (2b ± squrt(4) * squrt(b² - 2c + 1)
x = (2b ± 2 * squrt(b² - 2c + 1)

This is true because the product of square roots is equal to the
square root of the product.
See http://library.thinkquest.org/20991/alg/roots.html?tqskip1=1&tqtime=0905#Multiplication
for more information regarding this subject.
Hope that clears it up,
secret901-ga

Sorry, I suddenly realized that, but you had posted that response
already. The solution seems to work out for several values of a, b,
and c.

The Pythagorean theorem use (a² + b² - c² = 0) was rather clever.

Thanks for the swift response in both cases.

My apologies! The line:
a² + b² - 2bx + x = c² - 2c + 1 
should read:
a² + b² - 2bx + x² = c² - 2c + 1

The answer given is completely wrong.  The hypotenuse is never
shortened by 1 meter.  The size and shape of the right triangle formed
by the man, the sea, and the boat never changes.  If the man backs up
1 meter, the boat moves 1 meter closer, it's that simple.  You
describe a situation where the man stands still and pulls in the rope
which is not what was asked.

math_man:
The rope the man is holding onto forms the hypotenuse of the triangle.
 When he moves back 1 m, that's the same as standing still and pulling
the rope back by 1 m, because that 1 m will be a horizontal distance.
We assumed that 2 of the three endpoints forming the triangle are
fixed.

The Answer is still wrong and should be rejected.  No vertex of the
triangle is fixed, they all three move by 1 meter.  The boat is moved
closer to shore by one meter.   The boats' distance from the man stays
the same.  Standing still and pulling the rope is NOT!!!!!! the same
as BACKING UP!!!!!! by one meter.  A lot of effort was put into this
problem when it is basically a no-brainer.

math_man-ga, please stop arguing about the solution. Although I thank
you for trying to help, your argument is flawed. As I said in my
original question, one of the legs (vertical) is fixed (and as a
result, two vertices are fixed) because it's the height of the pier.

Further proof - the instructor showed a model of the situation, and
indeed, only two sides of the triangle (the hypotenuse and the
horizontal leg) moved.

If the accepted solution was incorrect, it is certainly not due to the
reasoning that you mentioned.

Sorry, but doesn't this depend on how long the rope is ? :)