Suppose there are three boats on a (flat) lake and one beach ball floating on the lake. At time t, it is given that the boats are equidistant from each other and that the ball is X meters from boat 1 and Y meters from boat 2. At that same point in time, what is the maximum distance the ball can be from boat 3?
Originally posted by davegage Suppose there are three boats on a (flat) lake and one beach ball floating on the lake. At time t, it is given that the boats are equidistant from each other and that the ball is X meters from boat 1 and Y meters from boat 2. At that same point in time, what is the maximum distance the ball can be from boat 3?
Originally posted by prosoccer sqrt (distance between boats)^2 + y^2
Do you mean sqrt of that whole expression?
That won't work. Here's a simple case that demonstrates why your expression cannot be an upper bound:
Suppose that the beach ball lies along an extension of the line connecting boat 3 to boat 2. Then the distance between the ball and boat three is equal to [(distance between boats) + y]. It should be fairly clear that this value is strictly greater than the value sqrt[(distance between boats)^2 + y^2].* This is a case which is consistent as a possibility with the information given, so your answer cannot be correct.
*Suppose the distance between boats is a. Then sqrt[a^2 + y^2] < sqrt[a^2 + 2ay + y^2] = sqrt[(a + y)^2] = a + y.
By the way, I'm looking for an answer that just includes specified distances x and y. The question only really makes sense in a context in which x and y are some fixed distances (that can constrain the possible size of the triangle formed by the boats).
I don't have time here at work to work through a complete solution, but I think this is the right place to start:
If the ball is (distance) X from (boat) 1, and Y from 2, then the maximum possible length of the sides of the equilateral triangle 1-2-3 is X+Y; ie. the ball is on the line 1-2.
The maximum distance the ball can possibly be from 3 in this instance is X+Y (when either X or Y = 0; at any other point on the line 1-2 it would be closer by simple Pythagoras).
If the length of the sides is not X+Y (ie. the ball does not lie on the line between 1-2) then the maximum distance the ball can be from 3 is:
(length of side of triangle) + (smaller of X and Y)
ie on the extension of the line 3-2, or 3-1 as applicable. The distance is known from the other point, putting the ball on a circle radius X (or Y) from boat 1 (or 2), and any other point on that circle is closer to 3.
In this case, the longer of X or Y must be longer than the lengths of the triangle sides, hence the distance is still less than X+Y.
So, to conclude from my appallingly worded/explained answer, I believe the answer is
Originally posted by Zeddicus I don't have time here at work to work through a complete solution, but I think this is the right place to start:
If the ball is (distance) X from (boat) 1, and Y from 2, then the maximum possible length of the sides of the equilateral triangle 1-2-3 is X+Y; ie. the ball is on the line 1-2.
The maximum distance the ball can possibly be from 3 in this ...[text shortened]... de from my appallingly worded/explained answer, I believe the answer is
Max Distance = X + Y
I very much like your reasoning here, and I agree with you answer. Good work.
However, I have one more question for you to think about. Your solution seems to demonstrate that X + Y is an upper bound, but other than trivial cases (i.e., where either Y = 0 or X = 0), it doesn't seem to demonstrate that it is achievable (i.e., cases where the length between boat 3 and the ball equals X + Y, not just where it is less than X + Y).
So my follow-up question would be can you specify a non-trivial case where the distance between boat 3 and the ball exactly equals the maximum possible, X + Y?
04 Apr '07 22:10
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