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riddles >> putnam exam (pure math) >> Altitudes of Triangle
(Message started by: ThudanBlunder on Jul 28th, 2009, 12:42am)

Title: Altitudes of Triangle
Post by ThudanBlunder on Jul 28th, 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 28th, 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 28th, 2009, 1:11am

on 07/28/09 at 00:45:35, 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?  ::)

Title: Re: Altitudes of Triangle
Post by Grimbal on Jul 28th, 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 28th, 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-1http://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.gifuv  f(csc t - sec t, sec t ) dt + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifvhttp://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.gifatan aatan 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+a2} - 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{1+b2}

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 28th, 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 x-y 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 28th, 2009, 6:56am
Ha, I knew that you (or another math whizz) would quickly see right through this 'problem', Eigenray.  ;)



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