Locus of the Incenter

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Let's first observe one of the right triangles.

Note that $AI$ and $BI$ are angle bisectors because of the definition of incenter. Thus, $\angle IAC = \angle IAB$ and $\angle IBC = \angle IBA$ .

Note that $\angle CAB + \angle CBA = 90^{\circ}$ , so $\angle IAB + \angle IBC = \dfrac{90^{\circ}}{2}=45^{\circ}$ .

Thus, $\angle AIB = 180^{\circ}-45^{\circ} = 135^{\circ}$ .

This means the locus of all possible incenters above the hypotenuse is a $360-135\times 2=90^{\circ}$ section of a circle:

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The mirror image on the other side forms the complete locus:

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We just need to find the area of the shaded region.

First, the side length of the square shown above is $\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$ . Thus, the area of it is $\dfrac{1}{2}$ .

Also, the area of the quarter section of one of the big circles is $\dfrac{1}{4}\pi\left(\dfrac{\sqrt{2}}{2}\right)^2=\dfrac{\pi}{8}$ . Thus, the area of both of them combined is $\dfrac{\pi}{4}$ .

Now we subtract the area of the square to cancel off the area that we don't need, and to get the area of the shaded region., Thus, $A=\dfrac{\pi}{4}-\dfrac{1}{2}\approx 0.285$

And our answer is $\boxed{285}$