Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
m (→‎Solution 2)
 
(7 intermediate revisions by 5 users not shown)
Line 8: Line 8:
 
<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.
  
−
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>.
+
-asdf334
  
−
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:
+
== 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>
−
<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>.)
 
  
 +
-jeteagle
  
 
== 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 12:25, 20 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 1

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


Solution 2

We multiply $z$ and $(\frac{3}{4}+\frac{3}{4}i)$ to get \[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\] Since we want to find the probability that this number is in $S$, we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives us the equations \[-1\le \frac{3}{4}x-\frac{3}{4}y \le 1\] and \[-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.\] Now, we see that we can solve this by graphing. We can graph our barriers $-1\le x\le 1$ and $-1\le y\le 1$ to form a $2$ by $2$ square centered at the origin. Graphing our two equations gives us the four lines \[x-y=\frac{4}{3},\] \[x-y=-\frac{4}{3},\] \[x+y=\frac{4}{3},\] \[x+y=-\frac{4}{3}.\] 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 $\frac{2}{9}$ area. Therefore, the octagon has area of $\frac{28}{9}.$ 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 $\frac{\frac{28}{9}}{4}=\frac{7}{9}$ or $\boxed{D}.$

-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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_23&oldid=115024"