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ThudnBlunder
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 Altitudes of Triangle   « on: Jul 28th, 2009, 12:42am » Quote Modify

What is the probability that the altitudes of a triangle may themselves form another triangle?
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Grimbal
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 Re: Altitudes of Triangle   « Reply #1 on: Jul 28th, 2009, 12:45am » Quote Modify

1.

The altitudes of a triangle may form another triangle.  I know at least one case.
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ThudnBlunder
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 Re: Altitudes of Triangle   « Reply #2 on: Jul 28th, 2009, 1:11am » Quote Modify

on Jul 28th, 2009, 12:45am, Grimbal wrote:
 1.   The altitudes of a triangle may form another triangle.  I know at least one case.

OK, what is the probability that they WILL form another triangle?
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Grimbal
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 Re: Altitudes of Triangle   « Reply #3 on: Jul 28th, 2009, 2:21am » Quote Modify

That depends on how you randomly draw a triangle, what is the distribution of triangles you consider.
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Eigenray
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 Re: Altitudes of Triangle   « Reply #4 on: Jul 28th, 2009, 3:33am » Quote Modify

For example, we could randomly pick a point on the unit sphere in the first octant; it will represent a triangle with probability
12/ cot-12  - 2 ~ 35.1%.
Conditioned on this point representing a triangle, the altitudes will form a triangle with probability ~ 58.1%.  But it's a nasty trig integral for the exact value: Let

A = uv  f(csc t - sec t, sec t ) dt + v/4  f(1/(sin t + cos t), sec t ) dt
where
u = sec-1 [ 5 / 2 ], v = tan-1 [ (5  - 1)/2 ],
and
f(a,b) = atan aatan b sin d= 1/{1+a2} - 1/{1+b2}

A ~ 0.0534 is the area of the region on the unit sphere satisfying
1/(1/y + 1/z) x y z x+y

Of course, a simpler approach is to set, say, z = 1, and compute the area of the set of x,y on the plane such that the above holds.  The set of (x,y) with 0 x y z x+y has area 1/4, so we multiply by 4 and get
2 - 5  + 4 log [ 5  - 1 ] ~ 61.2 %
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Eigenray
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 Re: Altitudes of Triangle   « Reply #5 on: Jul 28th, 2009, 4:15am » Quote Modify

Thirdly, we can pick a point on the plane x+y+z = 1; since this is linear the ratio of areas is the same if we project onto the x-y plane.  This obviously gives

[ 485  (log 2 - arccsch 2) + 455 - 110 ] / 25 ~ 53.5%

I believe that's my first time using the inverse of the hyperbolic cosecant  Well, that's what Mathematica gives.  We can also write
log 2 - arccsch 2 = log [ 5  - 1 ]
 « Last Edit: Jul 28th, 2009, 4:23am by Eigenray » IP Logged
ThudnBlunder
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 Re: Altitudes of Triangle   « Reply #6 on: Jul 28th, 2009, 6:56am » Quote Modify

Ha, I knew that you (or another math whizz) would quickly see right through this 'problem', Eigenray.
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