Question

A transformation of the first quadrant of the coordinate plane maps each point $(x, y)$ to the point $(\sqrt{x}, \sqrt{y}).$ The vertices of 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 $k.$

答案

Solution:

Follow the four edges. Sides $AB$ and $DC$ lie on the lines $y = \frac{x}{3}$ and $y = 3x,$ which map to the lines $v = \frac{u}{\sqrt{3}}$ and $v = \sqrt{3}\,u$ — rays from the origin at angles $30^\circ$ and $60^\circ.$ Sides $AD$ and $BC$ lie on $x + y = 1200$ and $x + y = 2400,$ which map to arcs of the circles $u^2 + v^2 = 1200$ and $u^2 + v^2 = 2400.$

So the image is the part of the annulus between radii $\sqrt{1200}$ and $\sqrt{2400}$ lying between the $30^\circ$ and $60^\circ$ rays, one twelfth of the full annulus: $k = \frac{30}{360}\,\pi\,(2400 - 1200) = 100\pi \approx 314.16.$

The greatest integer not exceeding $k$ is $314.$

1999 AIME Problem 6

Solution:

Problem 6 in Other Years