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I pick three numbers $(x_1, x_2, x_3)$, where the value each $x_i$ is a real number selected with uniform probability on the interval $[J, K]$. I then plot a point in three-dimensions where {$x, y, z$} = {$D(x_1, x_2)$, $D(x_2,x_3)$, $D(x_3,x_1)$}, where $D(a,b)$ is the straight-line distance between points $a$ and $b$. If I repeat this procedure a large number of times, what polyhedron's volume do my plotted points define?

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'Uniform probability' here is a red herring - the 'large number of times' implies that your result could be more easily gotten by looking at all tuples $(x_1,x_2,x_3)$.

As for the result, it's not actually a single polyhedron; in fact, it's a union of 3 planar figures. For concreteness, let's let $[J, K]$ be the unit interval and consider first the situation where $0\leq x_1\leq x_2 \leq x_3 \leq 1$. Then $u = D(x_1,x_2) = x_2-x_1$ can take all values in $[0,1]$, as can $v = D(x_2, x_3)$ and $w = D(x_3, x_1)$ - subject to the sole constraint that $u+v = w$. This implies that this 'fan' of your result is the triangle defined by the intersection of the plane $u+v=w$ with the unit cube $[0,1]^3$. (Exercise: convince yourself that this is an equilateral triangle!) Similarly, the situation where $0 \leq x_3 \leq x_2 \leq x_1 \leq 1$ spans exactly the same triangle (note that I'm treating distance as unsigned - if you mean the signed distance along the line then things change, but not as much as you might think; this is also a nice exercise!), since we know that the distance from $x_3$ to $x_1$ is the largest of the three and is thus the sum of the other two. The other possible orders of the $x$'s along the line correspond to the fans intersecting $u = v+w$ and $v = w+u$ with the unit cube, so your overall shape is the union of these three triangles.

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+1 These three triangles are three of the faces of a tetrahedron, made from the vertices $(0,0,0),\; (K-J,K-J,0),\; (K-J,0,K-J),\; (0,K-J,K-J)$, namely the three faces touching $(0,0,0)$. –  Henry Feb 7 '12 at 2:06

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