2009 AMC 12B Problems/Problem 23
Contents
Problem
A region 
in the complex plane is defined by
![\[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\]](http://latex.artofproblemsolving.com/1/f/1/1f183c6f855fb40cc0df24c6d24b6a99b1218b0d.png)
A complex number 
is chosen uniformly at random from . What is the probability that 
is also in ?
Solution 1
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 
degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.
-asdf334
Solution 2
We multiply 
and 
to get ![\[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\]](http://latex.artofproblemsolving.com/d/f/3/df372c2f4d1a3cec657afbf57fd48dd98c06d344.png)
Since we want to find the probability that this number is in , we need the real and complex coefficients of this number to be less than or equal to 
or greater than or equal to 
This gives us the equations ![\[-1\le \frac{3}{4}x-\frac{3}{4}y \le 1\]](http://latex.artofproblemsolving.com/8/f/5/8f5b87c094344b5099bc45d300a7fd2cf17cecda.png)
and ![\[-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.\]](http://latex.artofproblemsolving.com/6/4/1/6411ad90105848d1a572c97e7dde878d52cf6a73.png)
Now, we see that we can solve this by graphing. We can graph our barriers 
and 
to form a 
by square centered at the origin. Graphing our two equations gives us the four lines ![\[x-y=\frac{4}{3},\]](http://latex.artofproblemsolving.com/a/1/a/a1a1ec652ccebc89ff0f37eeb22f8ca04e85c1ec.png)
![\[x-y=-\frac{4}{3},\]](http://latex.artofproblemsolving.com/6/9/5/69564b0d4bf7bda9ba0a3eb6493b00d71953ae28.png)
![\[x+y=\frac{4}{3},\]](http://latex.artofproblemsolving.com/3/8/d/38d4d3f24ea42d139ecdd6e2c486aa578aed9648.png)
![\[x+y=-\frac{4}{3}.\]](http://latex.artofproblemsolving.com/8/f/b/8fbf7f5ef56ede769c975f5aa626e1df255f3067.png)
The square that is formed is the region that satisfies these four equations. Now, the barriers and this square gives us an octagon as the desired region. The area of this octagon is the total area of the square minus the 4 small triangles on each corner, each with 
area. Therefore, the octagon has area of 
Finally, to find the probability of it working, we find the area of the octagon divided by the area of the entire square which is 
or 
-jeteagle
See Also
| 2009 AMC 12B (Problems • Answer Key • 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 | |
| All AMC 12 Problems and Solutions | |
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