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Topic: Altitudes of Triangle (Read 1538 times) 

ThudnBlunder
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Altitudes of Triangle
« on: Jul 28^{th}, 2009, 12:42am » 
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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 28^{th}, 2009, 12:45am » 
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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 28^{th}, 2009, 1:11am » 
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on Jul 28^{th}, 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 28^{th}, 2009, 2:21am » 
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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 28^{th}, 2009, 3:33am » 
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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^{1}2  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 = _{u}^{v} 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 a}^{atan b} sin d= 1/{1+a^{2}}  1/{1+b^{2}} 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 28^{th}, 2009, 4:15am » 
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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 xy 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 28^{th}, 2009, 4:23am by Eigenray » 
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ThudnBlunder
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Re: Altitudes of Triangle
« Reply #6 on: Jul 28^{th}, 2009, 6:56am » 
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Ha, I knew that you (or another math whizz) would quickly see right through this 'problem', Eigenray.


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