Three trees

[BMAD]

Driving out in the "Western Plains" states of the U.S. is like being on a flat plane. Three trees are now growing at random points on the plane (or plain). What's the probability they form an obtuse triangle? Assume that two of the trees are a fixed distance 'x' from each other.

[vistaptb]

100%. If the trees are x's and at two fixed points (which we established earlier), then the situation looks like this:

l l
l l
x x
l l
l l
The third tree must be between the lines to make an acute triangle and anywhere else to make an obtuse triangle. Thus, the probability of an obtuse triangle being formed is 100%.

[bonanova]

Let y = the dimension of the Western plains: about 800 miles EW by 1400 miles NS.

Then acute probability p = x/y. where x is expressed in miles.

If the two trees are on an EW line, acute probability is p = x/800.

If the two trees are on a NS line, acute probability is p = x/1400.

If the two trees have a different orientation, p has an intermediate value.

If every case, obtuse probability is 1-p.

[jhawk]

Let the points be labeled A,B,C, with A & B, being the set points x distavs apart. Now we look at 2 different possibliites of length AC

.
If AC
Now consider AC > AB, As in length of AC increases in relation to AB, the probability of an obtuse triangle proportinately increases, with P approaching 1 as AC /AB approaches infinity.

Hmmm!, Now that I type this, I consider what happens when AB approaches infinity, Leaving P=0 for an obtuse triangle, Now I am totally lost and think it is unsolvable Or maybe it is just 50%.

,>

Edited April 17, 2013 by jhawk

[jhawk]

As the distace between any 2 trees inceases( to infinity?), the probability approaches 1. The universe may be infinite, but the plain is not. So it cannot be 100%, but some function of AC/AB. Lrt me think some more