A transformation of the first [quadrant](https://artofproblemsolving.com/wiki/index.php/Quadrant) of the [coordinate plane](https://artofproblemsolving.com/wiki/index.php/Coordinate_plane) maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$  The [vertices](https://artofproblemsolving.com/wiki/index.php/Vertex) of [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$  Let $k_{}$ be the area of the region enclosed by the image of quadrilateral $ABCD.$  Find the greatest integer that does not exceed $k_{}.$
\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ B' = & (\sqrt {1800}, \sqrt {600})\\ C' = & (\sqrt {600}, \sqrt {1800})\\ D' = & (\sqrt {300}, \sqrt {900}) \end{eqnarray*}
First we see that lines passing through $AB$ and $CD$ have [equations](https://artofproblemsolving.com/wiki/index.php/Equation) $y = \frac {1}{3}x$ and $y = 3x$, respectively. Looking at the points above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \frac {1}{3}x^2$ and $y^2 = 3x^2$, or, after manipulation $y = \frac {x}{\sqrt {3}}$ and $y = \sqrt {3}x$, respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines. 
Now take a look at $BC$ and $AD$, which have the equations $y = - x + 2400$ and $y = - x + 1200$. The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$, respectively, which are the equations for [circles](https://artofproblemsolving.com/wiki/index.php/Circle). 
[1999 AIME-6.png](https://artofproblemsolving.com/wiki/index.php/File:1999_AIME-6.png)
To find the area between the circles (actually, parts of the circles), we need to figure out the [angle](https://artofproblemsolving.com/wiki/index.php/Angle) of the [arc](https://artofproblemsolving.com/wiki/index.php/Arc). This could be done by $\arctan \sqrt {3} - \arctan \frac {1}{\sqrt {3}} = 60^\circ - 30^\circ = 30^\circ$. So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\frac {30^\circ}{360^\circ}(R^2\pi - r^2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi$. Hence the answer is $\boxed{314}$.