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        riddles >> putnam exam (pure math) >> Expected Area of Triangle
        
(Message started by: THUDandBLUNDER on Jan 2^nd, 2007, 10:51am)

Title: Expected Area of Triangle
Post by THUDandBLUNDER on Jan 2^nd, 2007, 10:51am

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?

Title: Re: Expected Area of Triangle
Post by Miles on Jan 3^rd, 2007, 2:34am

[hide]3 / 2pi[/hide]

[hideb]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). [/hideb]

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.

Title: Re: Expected Area of Triangle
Post by Miles on Jan 3^rd, 2007, 2:37am

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.

Title: Re: Expected Area of Triangle
Post by towr on Jan 3^rd, 2007, 2:54am

Do you mean [hide]3/2 pi[/hide] or [hide]3/(2 pi)[/hide]?
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.

Title: Re: Expected Area of Triangle
Post by Miles on Jan 3^rd, 2007, 3:01am

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).

Title: Re: Expected Area of Triangle
Post by Miles on Jan 3^rd, 2007, 5:50am

Come on then, what about 4 points round the circle? or N points?!

Title: Re: Expected Area of Triangle
Post by balakrishnan on Jan 11^th, 2007, 9:33am

For k points it is
A(k)=-k(k-1)/[2^k *pi^(k-1)] * I(k-2)

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^2-6)/(4*pi^3)
and so on

Title: Re: Expected Area of Triangle
Post by balakrishnan on Jan 11^th, 2007, 3:44pm

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*pi-x]/[Area of the hyperplane t1+..tk=2*pi]

Note that the numerator is proportional to (2*pi-x)^(k-2) ..since there are k-2 free variables,
while the denominator is proportional to
int[(2*pi-x)^(k-2)]_{x=0}^{2*pi} which is (2*pi)^(k-1)/(k-1)
and hence
and hence
Pr(x<=t1<=x+dx)=dx*(k-1)*[2*pi-x]^(k-2)/(2*pi)^(k-1)
and hence
E[Sin(t1)]=integral[Sin(x)*(k-1)*[2*pi-x]^(k-2)/(2*pi)^(k-1)*dx],{x=0 to 2*pi}
and hence our required expected area is simply
k/2*E[sin(t1)]
=
-k(k-1)/[2^k *pi^(k-1)] *integral[Sin(x)* x^(k-2)]_{x=0 to 2*pi}