2009 AMC 12B Problems/Problem 23
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Problem
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 ?
Solution
We can directly compute 
.
This number is in if and only if 
and at the same time 
. This simplifies to 
and 
.
Let 
, and let ![[X]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/7/a/a7a4eadb159b3eca3ae63991cbcf8a85bb2cffa5.gif)
denote the area of the region 
. Then obviously the probability we seek is ![\frac {[S\cap T]}{[S]} = \frac{[S\cap T]}4](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/d/a/d/dad49d24f6be0ff4ba9b8e2a210ede8c76e6616a.gif)
. 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 ![\frac{ [S\cap T]}4 = \frac{ 4 - \frac 89 }4 = 1 - \frac 29 = \boxed{\frac 79}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/7/9/679594597ed2703ded9ba90b3f7fdaf17c789f53.gif)
.
(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 
.)
See Also
| 2009 AMC 12B (Problems • Resources) | ||
| Preceded by Problem 22 | Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
