- 04 Apr '07 22:10Suppose 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?
- 05 Apr '07 02:29

sqrt (distance between boats)^2 + y^2*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?** - 05 Apr '07 07:47 / 3 edits

Do you mean sqrt of that whole expression?*Originally posted by prosoccer***sqrt (distance between boats)^2 + y^2**

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). - 05 Apr '07 13:16 / 1 editI 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 - 05 Apr '07 18:46

I very much like your reasoning here, and I agree with you answer. Good work.*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

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?