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

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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>
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A complex number <math>z = x + iy</math> is chosen uniformly at random fromm <math>S</math>. What is the probability that <math>\left(\frac34 + \frac34i\right)z</math> is also in <math>S</math>?
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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>

Revision as of 12:54, 2 August 2009

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

We can directly compute $\left(\frac34 + \frac34i\right)z = \left(\frac34 + \frac34i\right)(x + iy) = \frac{3(x-y)}4 + \frac{3(x+y)}4 \cdot i$.

This number is in $S$ if and only if $-1 \leq \frac{3(x-y)}4 \leq 1$ and at the same time $-1 \leq \frac{3(x+y)}4 \leq 1$. This simplifies to $|x-y|\leq\frac 43$ and $|x+y|\leq\frac 43$.

Let $T = \{ x + iy : |x-y|\leq\frac 43 ~\land~ |x+y|\leq\frac 43 \}$, and let $[X]$ denote the area of the region $X$. Then obviously the probability we seek is $\frac {[S\cap T]}{[S]} = \frac{[S\cap T]}4$. All we need to do is to compute the area of the intersection of $S$ and $T$. 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, $S$ is shown in red, $T$ in green and their intersection is yellow. The intersections of the boundary of $S$ and $T$ are obviously at $(\pm 1,\pm 1/3)$ and at $(\pm 1/3,\pm 1)$.

Hence each of the four red triangles is an isosceles right triangle with legs long $\frac 23$, and hence the area of a single red triangle is $\frac 12 \cdot \left( \frac 23 \right)^2 = \frac 29$. Then the area of all four is $\frac 89$, and therefore the area of $S\cap T$ is $4 - \frac 89$. Then the probability we seek is $\frac{ [S\cap T]}4 = \frac{ 4 - \frac 89 }4 = 1 - \frac 29 = \boxed{\frac 79}$.

(Alternately, when we got to the point that we know that a single red triangle is $\frac 29$, we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is $1 - \frac 29 = \frac 79$. This saves us the work of first multiplying and then dividing by $4$.)


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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