Problem 244

Consider the triangle bound by the x-axis, the y-axis, and the line $x+y=2$. The point equidistant from the vertices of this triangle is the incenter, the point of intersection of the angle bisectors and the center of the inscribed circle. Now, remove the coordinate system. Let the origin be $O$, the y-intercept of the line be $A$, the x-intercept of the line be $B$, and the point be $P$.

Notice that $x$ in the diagram is what we are looking for: the distance from the point to the x-axis ($OB$). Also, $OP, BP,$ and $AP$ are angle bisectors since $P$ is the incenter. $OPC\cong OPD$ by $AAS$, and $PD=OC$, since $PC||OD$, so $OC=CP=PD=DO=x$. Therefore, since $OA=OB=2$, we have $DA=CB=2-x$. Also, $CPB\cong FPB$ and $ADP\cong AFP$ by $AAS$, so $AF=FB=DA=CB=2-x$, and $AB=4-2x$. However, we know from the Pythagorean Theorem that $AB=2\sqrt{2}$. Therefore, $4-2x=2\sqrt{2}\implies x=2-\sqrt{2}, \boxed{\text{C}}$.