2009 AMC 12B Problems/Problem 23
A region in the complex plane is defined by A complex number is chosen uniformly at random from . What is the probability that is also in ?
We can directly compute .
This number is in if and only if and at the same time . This simplifies to and .
Let , and let denote the area of the region . Then obviously the probability we seek is . All we need to do is to compute the area of the intersection of and . It is easiest to do this graphically:
Coordinate axes are dashed, is shown in red, in green and their intersection is yellow. The intersections of the boundary of and are obviously at and at .
Hence each of the four red triangles is an isosceles right triangle with legs long , and hence the area of a single red triangle is . Then the area of all four is , and therefore the area of is . Then the probability we seek is .
(Alternately, when we got to the point that we know that a single red triangle is , we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is . This saves us the work of first multiplying and then dividing by .)
Solution 2 (Same idea)
The solution proposed above is good, but there is a more straightforward method. First, turn into polar form as . Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of and rotated $45\degree$ (Error compiling LaTeX. ! Undefined control sequence.)
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