No way out

The parabolas intersect at $(1,1)$ , so the coordinates of the inscribed circle will be $(1,r)$ .

Now look at the point $(a,a^{2})$ on the curve $y = x^{2}$ such that the normal to the curve at this point passes through $(1,r)$ . (By symmetry we only need to look at the "relationship" between one of the parabolas and the inscribed circle.) Since the slope of the curve at this point is $2a$ , the equation of the normal will be

$y - a^{2} = (-\frac{1}{2a})(x - a)$ ,

and since this will pass through $(1,r)$ we have that

$r - a^{2} = (-\frac{1}{2a})(1 - a)$ , (i).

Also, we know that the distance between $(a,a^{2})$ and $(1,r)$ is $r$ , and so

$(1 - a)^{2} + (r - a^{2})^{2} = r^{2}$ , (ii).

Combining equations (i) and (ii) we have that

$(1 - a)^{2} + (-\frac{1}{4a^{2}})(1 - a)^{2} = (a^{2} - (\frac{1}{2a})(1 - a))^{2}$

$\Longrightarrow (1 - a)^{2} = a^{4} - a + a^{2} \Longrightarrow a^{4} + a - 1 = 0$ ,

which by numerical methods has the positive real solution $a = 0.724491959...$ , which, when plugged into equation (i), yields the value $r = 0.334751...$ .

Thus $\lfloor 10000*r \rfloor = \boxed{3347}$ .

Assume a is radius of inscribed circle, it is easy to see that the centre of this circle on the line x=1: $(x-1)^2+(y-a)^2=a^2$ $(x-1)^2=a^2-(y-a)^2$ $y=\sqrt(a^2-(x-1)^2)+a$ $\sqrt{a^2-(x-1)^2}+a<x^2$ $\sqrt{a^2-(x-1)^2}<x^2-a$ $(a^2-(x-1)^2)<(x^2-a)^2$ $a^2-(x-1)^2<x^4-2ax^2+a^2$ $x^4-2ax^2+(x-1)^2>0$ $a<(x^4+(x-1)^2)/(2x^2)$ $=(1/2)(x^2+1-2/x+1/x^2$ $f(x)=x^2+1-2/x+1/x^2$ $f'=2*x+2/x^2-2/x^3$ f(x) changes its sign at the point has x-cord is: $x=0.725; f'>0; f_{min}=0,334751, \lfloor 10000*r \rfloor = \boxed{3347}$