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Difference between revisions of "2010 AMC 12B Problems/Problem 18"

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(Undo revision 44374 by Jhggins (talk))
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== Solution ==
 
== Solution ==
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===Solution 1===
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We will let the moves be complex numbers <math> a</math>, <math> b</math>, and <math> c</math>, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
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<cmath> \{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}</cmath>
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has magnitude less than or equal to <math> 1</math>. Hence, the probability is <math> \boxed{\text{(C)} \frac {1}{4}}</math>.
  
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The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in, the remaining places that the frog can jump to is a circle of radius 2 from where he landed. Every point in the circle of radius 2 is equally likely to be reached in two jumps.  The entirety of the circle you want to land in is enclosed in this larger circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.
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===Solution 2===
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The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in.  The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps.  The entirety of the circle you want to land in is enclosed in this larger disk, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}
 
{{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}

Revision as of 07:51, 26 January 2012

Problem

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

$\textbf{(A)}\ \dfrac{1}{6} \qquad \textbf{(B)}\ \dfrac{1}{5} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{1}{2}$

Solution

Solution 1

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set \[\{|a + b + c|, |a + b - c|, |a - b + c|, |a - b - c|\}\] has magnitude less than or equal to $1$. Hence, the probability is $\boxed{\text{(C)} \frac {1}{4}}$.

Solution 2

The first frog hop doesn't matter because no matter where you hop you are on the border of the circle you want to end in. The remaining places that the frog can jump to form a disk of radius 2 centered at the spot on which the frog first landed, and every point in the disk of radius 2 is equally likely to be reached in two jumps. The entirety of the circle you want to land in is enclosed in this larger disk, so you find the ratio of the two areas, which is $\boxed{\text{(C)} \frac {1}{4}}$.

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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