2013 AMC 8 Problems/Problem 21

Problem

Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution

$[asy] unitsize(8mm); for(int i=0; i<=8; ++i) { draw((0,i)--(8,i)); draw((i,0)--(i,8)); } fill((2,1)--(6,1)--(6,6)--(2,6)--cycle); for(int j=0; j<= 38; ++j) { draw((0,0)--(2,0)--(2,1)--(0,1)--(0,0), black+linewidth(3)); draw((6,6)--(6,8)--(8,8)--(8,6)--(6,6), black+linewidth(3)); }[/asy]$

We see that Samantha only has one way to go through in the city park, 6 ways to the city park to school, and 3 ways from her house to the city park. After this, we can see that there are $6 \cdot 3 = 18$ ways for Samantha to go to school. Hence, $\boxed{\textbf{E}}$

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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2013 AMC 8 Problems/Problem 21

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