1999 AIME Problem 6

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.

Answer

Answer: 314

Concepts:transformation coordinate geometry sector annulus

Difficulty rating: 2450

Small hint:

Instead of tracking the vertices, find what the map does to each of the four boundary lines

Big hint:

y = 3x

and

y = \frac{x}{3}

map to lines through the origin, while

x + y = c

maps to an arc of

u^2 + v^2 = c

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:

\begin{aligned} k &= \frac{30}{360}\,\pi\,(2400 - 1200) \\ &= 100\pi. \end{aligned}

Since

3.14 < \pi < 3.15,

we have

314 < k < 315,

so the greatest integer not exceeding

k

is

314.