1. Joined
    04 Aug '01
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    2408
    04 Apr '07 22:10
    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?
  2. Joined
    01 Apr '05
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    8760
    05 Apr '07 02:29
    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?
    sqrt (distance between boats)^2 + y^2
  3. Joined
    04 Aug '01
    Moves
    2408
    05 Apr '07 07:473 edits
    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).
  4. Joined
    12 Jul '06
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    1089
    05 Apr '07 13:161 edit
    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

    Max Distance = X + Y
  5. Joined
    04 Aug '01
    Moves
    2408
    05 Apr '07 18:46
    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?
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