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

Revision as of 22:41, 25 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 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 circle, so you find the ratio of the two areas, which is $\boxed{\text{(C)} \frac {1}{4}}$ .

@@ Line 12: / Line 12: @@
 ===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 is a circle of radius two 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 circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.
+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 circle, so you find the ratio of the two areas, which is <math>\boxed{\text{(C)} \frac {1}{4}}</math>.
 == See also ==
 {{AMC12 box|year=2010|num-b=17|num-a=19|ab=B}}

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