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A point P is randomly chosen on the triangle with sides' length 1. The triangle is spun randomly (uniformly) about its vertex (0,0). Let (X, Y) denote P's coordinate. Find the joint density of (X, Y).

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In polar coordinates $(R,\Theta)$, obviously $\Theta$ is uniformly distributed on $[0,2\pi)$, and $R$ is independent of $\Theta$. We only need to know the distribution of $R$. Two of the sides of the triangle give uniform distributions on $[0,1]$. I'll let you do the other one.

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Thanks for the hint! $f(x,y)$ over the two sides of the triangle could be easily obtained by following the "standard" procedure. It turns out each of whose is $$ f(x,y) = \frac{\sqrt{x^2+y^2}}{2\pi}$$. For the 3rd side, the following relationship bould be obtain by some geometry: $$ X\ =\ \cos(\frac{\pi}{3} + \theta) + t \cos(\frac{\pi}{3} - \theta) \\ Y\ =\ \sin(\frac{\pi}{3} + \theta) + t \sin(\frac{\pi}{3} - \theta)$$. Where $\theta \in [0,2\pi]$, and $t \in [0,1]$. – Jimmy7430 Jul 29 '12 at 4:09
Solving the Jacobian for the 3rd edge seems infeasible. In order to have some intuition, I also plot the parametric graph of (x,y): It seems that transformation is not one-to-one over the 3rd edge, since the Sharp Pointers were missing from the 2nd and 4th Quadrant. – Jimmy7430 Jul 29 '12 at 4:27
The closest point to the origin on the third side is at distance $\sqrt{3}/2$. For $\sqrt{3}/2 < r < 1$, the part of the third side with distance $<r$ from the origin has length $2 \sqrt{r^2 - 3/4}$ – Robert Israel Jul 29 '12 at 6:31

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