More - take it to either extreme - if he's level with the boat it will advance one yard for one yard of string. If he's almost above it, then the difference in rope-length between the boat being a few feet from shore and aground will be next to nothing.
Unfortunately, the answer and explanation at the link is wrong. Sure, b > a-c, but that equation does not determine the absolute relationship of b and a. If you consider the starting equation of b+c>a then you notice that you must add something to b in order to exceed a -- so a must be the greater number. (To be precise, these equations should be >= for the case of c=0.)
The correct answer is less (or equal if c=0). Consider the extreme case where the boy is on the shore (c=0), then the rope and the distance traveled are the same (a=b). If you stand above the shore (c>0), then it takes a longer rope to reach the object in the water. Thus, when you pull in 1m of rope, the boat will move some smaller distance.
As an English major, I don't mind sounding stupid re math, so I'll ask what may be a dumb question. I don't see any need for the equation to "determine the absolute relationship of b and a." It doesn't seem to matter if the ratio is 1:1.5 or 1:10.
And I agree that when the puller is up high, it "takes a longer rope to reach the object", but it doesn't logically follow that "when you pull in 1m of rope, the boat will move some smaller distance. Why can't it move more than a yard?
Too rapid a post... The author is mostly correct. I got stuck on the special case of c=0 in which case the changes are the same. However, when c is greater than 0 the boat does move faster.
I too have a problem with this. First, the problem remains the same if the "C" side is at right angles to the shore line, making this a a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
In the original condition, we have b*b = a*a - c*c, and in the second case, we have b*b = (a-1)*(a-1) - b'*b'.
Because b remains constant, we have a*a - c*c = (a-1)*(a-1) - b'*b'.
Expanding, we have a*a - c*c = a*a -2a + 1 - b'*b' or -c*c = -2a + 1 - b'*b' or 2a - c*c - 1 = -b'*b' or c*c - 2a + 1 = b'*b'
Completing the square, we have
c*c - 2c + 1 - 2a + 2c = b'*b' or (c-1)*(c-1) - 2(a + c) = b'*b'
This means that b' is < c-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
Argh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!
I too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.
Because c remains constant, we have a*a - b*b = (a-1)*(a-1) - b'*b'.
Expanding, we have a*a - b*b = a*a - 2a + 1 - b'*b' or -b*b = -2a + 1 - b'*b' or 2a - b*b - 1 = -b'*b' or b*b - 2a + 1 = b'*b'
Completing the square, we have
b*b - 2b + 1 - 2a + 2b = b'*b' or (b-1)*(b-1) - 2(a + b) = b'*b'
This means that b' is < b-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
One more time. Need to get more than 2 hours of sleep a night. I mashed a sign near the end:
Argh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!
I too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.
Because c remains constant, we have a*a - b*b = (a-1)*(a-1) - b'*b'.
Expanding, we have a*a - b*b = a*a - 2a + 1 - b'*b' or -b*b = -2a + 1 - b'*b' or 2a - b*b - 1 = -b'*b' or b*b - 2a + 1 = b'*b'
Completing the square, we have
b*b - 2b + 1 - 2a + 2b = b'*b' or (b-1)*(b-1) - 2(a - b) = b'*b'
This means that (because a > b and a-b is positive), that b' is < b-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
Say the stick figure is 15 feet above the water (c), the toy is 36 feet out (b), and the string is 39 feet long (a). Those are the proportions of a 5-12-13 triangle, with a right angle below his feet.
He then pulls until the triangle becomes a 3-4-5 right triangle. He is still 15 feet up (c). That doesn't change. The boat is 20 feet out (b), and the string is 25 feet long (a).
He has pulled in 39-25 = 14 feet of string, and the boat has moved 36-20 = 16 feet.
The example proves the boat CAN move more than the length of the string pulled; it doesn't per se prove it always moves the greater distance.
I've finally figured out why this problem's solution is so counterintuitive. When we hear, "pull in one unit on the hypotenuse," the thought immediately is "similar triangles!" in which case the boat will approach the shore at less than one unit.
BUT: These are NOT similar triangles; the angle at the point of pull, relative to vertical, becomes smaller. The boat is then much like a pendulum approaching its own vertical.
Just from fishing experience and not bothering with math, the bobber reels in faster as it gets closer to the rod. It's faster still when you're on an elevated pier. So the solution is pretty intuitive for me.
(A) One yard pulled in equals one yard of boat distance reduction. Unless you have some rope that elastically stretches out when pulled or magically shrinks when pulled.
This isn't a trigonometry problem. It is basic, pulling rope 1 foot results in object at the end of the rope getting nearer to you by 1 foot problem.
Here is a similar question with the above diagram. The boy is holding a long thin semi-rigid metal rod. The end of the rod is 10 feet away. The boy while holding the rod steps 1 foot backwards, will the end of the rod also move 1 foot closer to the shore? This isn't a trigonometry problem, it is a low-grade intelligence test.
More - take it to either extreme - if he's level with the boat it will advance one yard for one yard of string. If he's almost above it, then the difference in rope-length between the boat being a few feet from shore and aground will be next to nothing.
ReplyDeleteUnfortunately, the answer and explanation at the link is wrong. Sure, b > a-c, but that equation does not determine the absolute relationship of b and a. If you consider the starting equation of b+c>a then you notice that you must add something to b in order to exceed a -- so a must be the greater number. (To be precise, these equations should be >= for the case of c=0.)
ReplyDeleteThe correct answer is less (or equal if c=0). Consider the extreme case where the boy is on the shore (c=0), then the rope and the distance traveled are the same (a=b). If you stand above the shore (c>0), then it takes a longer rope to reach the object in the water. Thus, when you pull in 1m of rope, the boat will move some smaller distance.
As an English major, I don't mind sounding stupid re math, so I'll ask what may be a dumb question. I don't see any need for the equation to "determine the absolute relationship of b and a." It doesn't seem to matter if the ratio is 1:1.5 or 1:10.
ReplyDeleteAnd I agree that when the puller is up high, it "takes a longer rope to reach the object", but it doesn't logically follow that "when you pull in 1m of rope, the boat will move some smaller distance. Why can't it move more than a yard?
This comment has been removed by the author.
ReplyDeleteToo rapid a post... The author is mostly correct. I got stuck on the special case of c=0 in which case the changes are the same. However, when c is greater than 0 the boat does move faster.
ReplyDeleteI too have a problem with this. First, the problem remains the same if the "C" side is at right angles to the shore line, making this a a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
ReplyDeleteIn the original condition, we have b*b = a*a - c*c, and in the second case, we have b*b = (a-1)*(a-1) - b'*b'.
Because b remains constant, we have a*a - c*c = (a-1)*(a-1) - b'*b'.
Expanding, we have
a*a - c*c = a*a -2a + 1 - b'*b'
or
-c*c = -2a + 1 - b'*b'
or
2a - c*c - 1 = -b'*b'
or
c*c - 2a + 1 = b'*b'
Completing the square, we have
c*c - 2c + 1 - 2a + 2c = b'*b'
or
(c-1)*(c-1) - 2(a + c) = b'*b'
This means that b' is < c-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
Lurker111
Argh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!
ReplyDeleteI too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.
Because c remains constant, we have a*a - b*b = (a-1)*(a-1) - b'*b'.
Expanding, we have
a*a - b*b = a*a - 2a + 1 - b'*b'
or
-b*b = -2a + 1 - b'*b'
or
2a - b*b - 1 = -b'*b'
or
b*b - 2a + 1 = b'*b'
Completing the square, we have
b*b - 2b + 1 - 2a + 2b = b'*b'
or
(b-1)*(b-1) - 2(a + b) = b'*b'
This means that b' is < b-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
Lurker111
One more time. Need to get more than 2 hours of sleep a night. I mashed a sign near the end:
ReplyDeleteArgh. I changed some variable names from my paper calculation, and messed up the post. Let me try again!
I too have a problem with this. First, the problem remains the same if the "c" side is at right angles to the shore line, making this a right triangle. Then, a*a = b*b + c*c. The question then reduces to: If a is reduced to a-1, and c remains constant, what is the value of the new b, call it b'? Or, (a-1)*(a-1) = b'*b' + c*c.
In the original condition, we have c*c = a*a - b*b, and in the second case, we have c*c = (a-1)*(a-1) - b'*b'.
Because c remains constant, we have a*a - b*b = (a-1)*(a-1) - b'*b'.
Expanding, we have
a*a - b*b = a*a - 2a + 1 - b'*b'
or
-b*b = -2a + 1 - b'*b'
or
2a - b*b - 1 = -b'*b'
or
b*b - 2a + 1 = b'*b'
Completing the square, we have
b*b - 2b + 1 - 2a + 2b = b'*b'
or
(b-1)*(b-1) - 2(a - b) = b'*b'
This means that (because a > b and a-b is positive), that b' is < b-1, and so the horizontal distance difference is less than the 1 unit removed from the hypotenuse.
Somebody please kill my two earlier posts! Sorry!
Lurker111
Argh. If b' is < b-1, then the horizontal difference IS greater than the one unit.
ReplyDeleteSomeone shoot me.
Lurker111
Here's a concrete example.
ReplyDeleteSay the stick figure is 15 feet above the water (c), the toy is 36 feet out (b), and the string is 39 feet long (a). Those are the proportions of a 5-12-13 triangle, with a right angle below his feet.
He then pulls until the triangle becomes a 3-4-5 right triangle. He is still 15 feet up (c). That doesn't change. The boat is 20 feet out (b), and the string is 25 feet long (a).
He has pulled in 39-25 = 14 feet of string, and the boat has moved 36-20 = 16 feet.
The example proves the boat CAN move more than the length of the string pulled; it doesn't per se prove it always moves the greater distance.
I've finally figured out why this problem's solution is so counterintuitive. When we hear, "pull in one unit on the hypotenuse," the thought immediately is "similar triangles!" in which case the boat will approach the shore at less than one unit.
ReplyDeleteBUT: These are NOT similar triangles; the angle at the point of pull, relative to vertical, becomes smaller. The boat is then much like a pendulum approaching its own vertical.
Lurker111
Just from fishing experience and not bothering with math, the bobber reels in faster as it gets closer to the rod. It's faster still when you're on an elevated pier. So the solution is pretty intuitive for me.
ReplyDelete(A) One yard pulled in equals one yard of boat distance reduction.
ReplyDeleteUnless you have some rope that elastically stretches out when pulled or magically shrinks when pulled.
This isn't a trigonometry problem. It is basic, pulling rope 1 foot results in object at the end of the rope getting nearer to you by 1 foot problem.
Here is a similar question with the above diagram.
The boy is holding a long thin semi-rigid metal rod. The end of the rod is 10 feet away. The boy while holding the rod steps 1 foot backwards, will the end of the rod also move 1 foot closer to the shore?
This isn't a trigonometry problem, it is a low-grade intelligence test.
"(A) One yard pulled in equals one yard of boat distance reduction"
DeleteBut... he's pulling diagonally, and the boat is moving horizontally (unless he's lifting it out of the water.
So one does not necessarily equal one.