

Title: Altitudes of Triangle Post by ThudanBlunder on Jul 28^{th}, 2009, 12:42am What is the probability that the altitudes of a triangle may themselves form another triangle? 

Title: Re: Altitudes of Triangle Post by Grimbal on Jul 28^{th}, 2009, 12:45am 1. The altitudes of a triangle may form another triangle. I know at least one case. 

Title: Re: Altitudes of Triangle Post by ThudanBlunder on Jul 28^{th}, 2009, 1:11am on 07/28/09 at 00:45:35, Grimbal wrote:
OK, what is the probability that they WILL form another triangle? ::) 

Title: Re: Altitudes of Triangle Post by Grimbal on Jul 28^{th}, 2009, 2:21am That depends on how you randomly draw a triangle, what is the distribution of triangles you consider. 

Title: Re: Altitudes of Triangle Post by Eigenray on Jul 28^{th}, 2009, 3:33am For example, we could randomly pick a point on the unit sphere in the first octant; it will represent a triangle with probability 12/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif cot^{1}http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2  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 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{u}^{v} f(csc t  sec t, sec t ) dt + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{v}^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4} f(1/(sin t + cos t), sec t ) dt where u = sec^{1} [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 / 2 ], v = tan^{1} [ (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5  1)/2 ], and f(a,b) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{atan a}^{atan b} sin http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif= 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{1+a^{2}}  1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{1+b^{2}} A ~ 0.0534 is the area of the region on the unit sphere satisfying 1/(1/y + 1/z) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif z http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif z http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif x+y has area 1/4, so we multiply by 4 and get 2  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 + 4 log [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5  1 ] ~ 61.2 % 

Title: Re: Altitudes of Triangle Post by Eigenray on Jul 28^{th}, 2009, 4:15am 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 [ 48http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 (log 2  arccsch 2) + 45http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5  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 [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5  1 ] 

Title: Re: Altitudes of Triangle Post by ThudanBlunder on Jul 28^{th}, 2009, 6:56am Ha, I knew that you (or another math whizz) would quickly see right through this 'problem', Eigenray. ;) 

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