2009 AMC 12B Problem 23

Question

A region

S

in the complex plane is defined by

\scriptsize S = \{x + iy : -1 \le x \le 1,\ -1 \le y \le 1\}.

A complex number

z = x + iy

is chosen uniformly at random from

S.

What is the probability that

\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)z

is also in

S?

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{7}{9}

\dfrac{7}{8}

Multiplying gives real part

\tfrac{3}{4}(x - y)

and imaginary part

\tfrac{3}{4}(x + y)

; both must lie in

[-1, 1]

These conditions are

|x - y| \le \tfrac{4}{3}

and

|x + y| \le \tfrac{4}{3}

; find the area of

S

satisfying them

Expanding,

\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy)

= \dfrac{3}{4}(x - y)

+ \dfrac{3}{4}(x + y)i.

Both parts lie in

[-1, 1]

iff

|x - y| \le \dfrac{4}{3}

and

|x + y| \le \dfrac{4}{3}.

Within the square

S

(area

4

) these fail only in four corner triangles. Near

(1, 1),

the line

x + y = \dfrac{4}{3}

cuts off a right triangle with legs

\dfrac{2}{3},

area

\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}.

The four corners remove

4 \cdot \dfrac{2}{9} = \dfrac{8}{9},

leaving

4 - \dfrac{8}{9} = \dfrac{28}{9}.

The probability is

\dfrac{\frac{28}{9}}{4} = \dfrac{7}{9}.

Thus, the correct answer is D.

Problem 23 in other years