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Topic: Expected Area of Triangle (Read 1026 times) 

ThudnBlunder
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Expected Area of Triangle
« on: Jan 2^{nd}, 2007, 10:51am » 
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If three points are randomly and uniformly chosen on the circumference of a circle of unit radius, what is the expected area of the resulting triangle?


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Miles
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Re: Expected Area of Triangle
« Reply #1 on: Jan 3^{rd}, 2007, 2:34am » 
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3 / 2pi hidden:  I got this by fixing one point and specifying the second point as being x radians round the circle from the first, and the third point y radians round the circle from the second. The area is then A = sin x + sin y + sin(2pi  x  y) / 2 and the expected area simplifies to integrating (1/4pi^2)*(sin x + sin y  sin(x+y)) over the region {0<y<2pi, 0<x<2pi  y} (I'm integrating where the expression is positive and then doubling).  Had to dredge up an old memory of how to integrate y*sin(y) from school nearly 20 years ago. [Edit (removed erroneous comment)] Anyway, I also agreed the above result with some numerical experimentation.

« Last Edit: Jan 3^{rd}, 2007, 2:47am by Miles » 
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Miles
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Re: Expected Area of Triangle
« Reply #2 on: Jan 3^{rd}, 2007, 2:37am » 
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And that post just earned me junior membership  appropriate that I got it on a maths problem as that's what brought me to this site in the first place.


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towr
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Re: Expected Area of Triangle
« Reply #3 on: Jan 3^{rd}, 2007, 2:54am » 
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Do you mean 3/2 pi or 3/(2 pi)? It's probably the latter for obvious reasons (because the first would be bigger than the circle), but as it's written now it's a bit ambiguous. And congratulations on the extra star.

« Last Edit: Jan 3^{rd}, 2007, 2:56am by towr » 
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Miles
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Re: Expected Area of Triangle
« Reply #4 on: Jan 3^{rd}, 2007, 3:01am » 
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The latter as you say. (If I had used the greek letter for pi instead of "pi" it wouldn't have been ambiguous under normal algebraic conventions).


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Miles
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Re: Expected Area of Triangle
« Reply #5 on: Jan 3^{rd}, 2007, 5:50am » 
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Come on then, what about 4 points round the circle? or N points?!


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balakrishnan
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Re: Expected Area of Triangle
« Reply #6 on: Jan 11^{th}, 2007, 9:33am » 
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For k points it is A(k)=k(k1)/[2^k *pi^(k1)] * I(k2) where I(m)=integral[x^m sin[x]]_{x=0 to 2*pi} eg for 3 points it si A(3)=3*2/(2^3*pi^2) * I(1) and I(1)=2*pi So A(3)=3/(2*pi) A(4)=3/pi A(5)=5*(4*pi^26)/(4*pi^3) and so on

« Last Edit: Jan 11^{th}, 2007, 11:40am by balakrishnan » 
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balakrishnan
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Re: Expected Area of Triangle
« Reply #7 on: Jan 11^{th}, 2007, 3:44pm » 
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Here is the derivation. If we have k points Let t1,t2...tk be the angle differences between the adjacent points. We have t1+t2+..+tk=2*pi Now the ti's are chosen uniformly on the hyperplane t1+t2+..+tk=2*pi Note that the area of the polygon is 1/2*[sin(t1)+sin(t2)+...+sin(tk)] So the expected value is simply k/2*E[sin(t1)] Since E[Sin(ti)]=E[Sin(tj)] as they are identical Now Let us compute the probability distribution of t1. for a small dx Pr(x<=t1<=x+dx)=dx*[Area of the hyperplane t2+t2+..t(k)=2*pix]/[Area of the hyperplane t1+..tk=2*pi] Note that the numerator is proportional to (2*pix)^(k2) ..since there are k2 free variables, while the denominator is proportional to int[(2*pix)^(k2)]_{x=0}^{2*pi} which is (2*pi)^(k1)/(k1) and hence and hence Pr(x<=t1<=x+dx)=dx*(k1)*[2*pix]^(k2)/(2*pi)^(k1) and hence E[Sin(t1)]=integral[Sin(x)*(k1)*[2*pix]^(k2)/(2*pi)^(k1)*dx],{x=0 to 2*pi} and hence our required expected area is simply k/2*E[sin(t1)] = k(k1)/[2^k *pi^(k1)] *integral[Sin(x)* x^(k2)]_{x=0 to 2*pi}

« Last Edit: Jan 11^{th}, 2007, 3:51pm by balakrishnan » 
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