Difference between revisions of "2009 AMC 12B Problems/Problem 23"

m (Solution 2 (Same idea))
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== Solution ==
 
== Solution ==
 
+
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.
We can directly compute <math>\left(\frac34 + \frac34i\right)z = \left(\frac34 + \frac34i\right)(x + iy) = \frac{3(x-y)}4 + \frac{3(x+y)}4 \cdot i</math>.
 
 
 
This number is in <math>S</math> if and only if <math>-1 \leq \frac{3(x-y)}4 \leq 1</math> and at the same time <math>-1 \leq \frac{3(x+y)}4 \leq 1</math>. This simplifies to <math>|x-y|\leq\frac 43</math> and <math>|x+y|\leq\frac 43</math>.
 
 
 
Let <math>T = \{ x + iy : |x-y|\leq\frac 43 ~\land~ |x+y|\leq\frac 43 \}</math>, and let <math>[X]</math> denote the area of the region <math>X</math>. Then obviously the probability we seek is <math>\frac {[S\cap T]}{[S]} = \frac{[S\cap T]}4</math>. All we need to do is to compute the area of the intersection of <math>S</math> and <math>T</math>. It is easiest to do this graphically:
 
 
 
<asy>
 
unitsize(2cm);
 
defaultpen(0.8);
 
path s = (-1,-1) -- (-1,1) -- (1,1) -- (1,-1) -- cycle;
 
path t = (4/3,0) -- (0,4/3) -- (-4/3,0) -- (0,-4/3) -- cycle;
 
path s_cap_t = (1/3,1) -- (1,1/3) -- (1,-1/3) -- (1/3,-1) -- (-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle;
 
filldraw(s, lightred, black);
 
filldraw(t, lightgreen, black);
 
filldraw(s_cap_t, lightyellow, black);
 
draw( (-5/3,0) -- (5/3,0), dashed );
 
draw( (0,-5/3) -- (0,5/3), dashed );
 
</asy>
 
 
 
Coordinate axes are dashed, <math>S</math> is shown in red, <math>T</math> in green and their intersection is yellow. The intersections of the boundary of <math>S</math> and <math>T</math> are obviously at <math>(\pm 1,\pm 1/3)</math> and at <math>(\pm 1/3,\pm 1)</math>.
 
 
 
Hence each of the four red triangles is an isosceles right triangle with legs long <math>\frac 23</math>, and hence the area of a single red triangle is <math>\frac 12 \cdot \left( \frac 23 \right)^2 = \frac 29</math>. Then the area of all four is <math>\frac 89</math>, and therefore the area of <math>S\cap T</math> is <math>4 - \frac 89</math>. Then the probability we seek is <math>\frac{ [S\cap T]}4 = \frac{ 4 - \frac 89 }4 = 1 - \frac 29 = \boxed{\frac 79}</math>.
 
 
 
(Alternately, when we got to the point that we know that a single red triangle is <math>\frac 29</math>, we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is <math>1 - \frac 29 = \frac 79</math>. This saves us the work of first multiplying and then dividing by <math>4</math>.)
 
 
 
== 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 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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Revision as of 07:22, 4 January 2020

Problem

A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$. What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$?

$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$

Solution

First, turn $\frac34 + \frac34i$ into polar form as $\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}$. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of $\frac{3\sqrt{2}}{4}$ and rotated $45$ degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.

-asdf334

See Also

2009 AMC 12B (ProblemsAnswer KeyResources)
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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