Question

A region $S$ in the complex plane is defined by $S = \{x + iy : -1 \le x \le 1,\ -1 \le y \le 1\}.$

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 $S?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{9}$

$\dfrac{7}{8}$

Solution:

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 $\dfrac{28/9}{4} = \dfrac{7}{9}.$

Thus, the correct answer is D.

2009 AMC 12B Problem 23

Solution:

Problem 23 in Other Years