2009 AMC 12B 第 23 题

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.