1969 AMC 12 Problem 31

Question

Let

OABC

be a unit square in the

xy

-plane with

O(0,0),

A(1,0),

B(1,1),

and

C(0,1).

Let

u=x^2-y^2

and

v=2xy

define a transformation of the

xy

-plane into the

uv

-plane. The transform (or image) of the square is:

Map the four vertices and then transform each side separately

The horizontal sides and vertical sides become two line segments and two parabolic arcs

The vertices map as

\begin{aligned} O&\mapsto(0,0),\\ A&\mapsto(1,0),\\ B&\mapsto(0,2),\\ C&\mapsto(-1,0). \end{aligned}

Side

OA

maps to the segment from

(0,0)

to

(1,0),

and

CO

maps to the segment from

(-1,0)

to

(0,0).

On

AB,

x=1,

so

u=1-\frac{v^2}{4};

on

BC,

y=1,

so

u=\frac{v^2}{4}-1.

These are the two upper parabolic arcs joining

(\pm1,0)

to

(0,2).

Therefore, the correct answer is D.

Problem 31 in other years