Difference between revisions of "2009 AMC 12B Problems/Problem 23"
(New page: == Problem == A region <math>S</math> in the complex plane is defined by <cmath> S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}. </cmath> A complex number <math>z = x + iy</math> is chosen uni...) |
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S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}. | S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}. | ||
</cmath> | </cmath> | ||
− | A complex number <math>z = x + iy</math> is chosen uniformly at random | + | A complex number <math>z = x + iy</math> is chosen uniformly at random from <math>S</math>. What is the probability that <math>\left(\frac34 + \frac34i\right)z</math> is also in <math>S</math>? |
<math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78</math> | <math>\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78</math> | ||
− | == Solution == | + | == Solution 1 == |
+ | 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. | ||
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− | + | == Solution 2 == | |
− | + | We multiply <math>z</math> and <math>(\frac{3}{4}+\frac{3}{4}i)</math> to get <cmath>(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).</cmath> Since we want to find the probability that this number is in <math>S</math>, we need the real and complex coefficients of this number to be less than or equal to <math>1</math> or greater than or equal to <math>-1.</math> This gives us the equations <cmath>-1\le \frac{3}{4}x-\frac{3}{4}y \le 1</cmath> and <cmath>-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.</cmath> Now, we see that we can solve this by graphing. We can graph our barriers <math>-1\le x\le 1</math> and <math>-1\le y\le 1</math> to form a <math>2</math> by <math>2</math> square centered at the origin. Graphing our two equations gives us the four lines <cmath>x-y=\frac{4}{3},</cmath> <cmath>x-y=-\frac{4}{3},</cmath> <cmath>x+y=\frac{4}{3},</cmath> <cmath>x+y=-\frac{4}{3}.</cmath> 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 <math>\frac{2}{9}</math> area. Therefore, the octagon has area of <math>\frac{28}{9}.</math> 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 <math>\frac{\frac{28}{9}}{4}=\frac{7}{9}</math> or <math>\boxed{D}.</math> | |
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== 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}} |
Latest revision as of 13:25, 20 January 2020
Contents
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 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
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
and
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.