Difference between revisions of "2009 AMC 12B Problems/Problem 23"
m (→Solution 2 (Same idea)) |
|||
Line 36: | Line 36: | ||
== Solution 2 (Same idea) == | == Solution 2 (Same idea) == | ||
− | The solution proposed above is good, but there is a more straightforward method. First, turn <math>\frac34 + \frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of <math>\frac{3\sqrt{2}}{4}</math> and rotated <math>45</math> degrees. This skips all the absolute values required before. Finish with the symmetry method stated above. | + | The solution proposed above is good, but there is a more straightforward method. First, turn <math>\frac34 + \frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of <math>\frac{3\sqrt{2}}{4}</math> and rotated <math>45</math> degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above. |
-Rowechen | -Rowechen | ||
+ | |||
== See Also == | == See Also == | ||
{{AMC12 box|year=2009|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2009|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:38, 12 December 2018
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 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 degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.
-Rowechen
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.