2009 AMC 12B Problem 23

Attempt Problem 23 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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23.

A region SS in the complex plane is defined by S={x+iy:1x1, 1y1}. \scriptsize S = \{x + iy : -1 \le x \le 1,\ -1 \le y \le 1\}.

A complex number z=x+iyz = x + iy is chosen uniformly at random from S.S. What is the probability that (34+34i)z\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)z is also in S?S?

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

79\dfrac{7}{9}

78\dfrac{7}{8}

Answer: D
Concepts:complex numbergeometric probabilityarea
Difficulty rating: 2340
Solution:

Expanding, (34+34i)(x+iy)\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy) =34(xy)= \dfrac{3}{4}(x - y) +34(x+y)i.+ \dfrac{3}{4}(x + y)i. Both parts lie in [1,1][-1, 1] iff xy43|x - y| \le \dfrac{4}{3} and x+y43.|x + y| \le \dfrac{4}{3}.

Within the square SS (area 44) these fail only in four corner triangles. Near (1,1),(1, 1), the line x+y=43x + y = \dfrac{4}{3} cuts off a right triangle with legs 23,\dfrac{2}{3}, area 122323=29.\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}.

The four corners remove 429=89,4 \cdot \dfrac{2}{9} = \dfrac{8}{9}, leaving 489=289.4 - \dfrac{8}{9} = \dfrac{28}{9}. The probability is 28/94=79.\dfrac{28/9}{4} = \dfrac{7}{9}.

Thus, the correct answer is D.

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