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Difference between revisions of "2011 AMC 12B Problems/Problem 23"

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==Solution==
 
==Solution==
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If a point <math>(x, y)</math> satisfies the property that <math>|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20</math>, then it is in the desirable range because <math>|x - 3| + |y + 2|</math> is the length of the shortest path from <math>(x,y)</math> to <math>B</math>, and <math>|x + 3| + |y - 2|</math> is the length of the shortest path from <math>(x,y)</math> to <math>A</math>
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We declare a point <math>(x, y)</math> to make up for the extra steps that the bug has to move. If the point <math>(x, y)</math> satisfies the property that <math>|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20</math>, then it is in the desirable range because <math>|x - 3| + |y + 2|</math> is the length of the shortest path from <math>(x,y)</math> to <math>(3, -2)</math>, and <math>|x + 3| + |y - 2|</math> is the length of the shortest path from <math>(x,y)</math> to <math>(-3, 2)</math>
  
  

Revision as of 21:33, 19 January 2020

Problem

A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$-axis or $y$-axis. Let $A = (-3, 2)$ and $B = (3, -2)$. Consider all possible paths of the bug from $A$ to $B$ of length at most $20$. How many points with integer coordinates lie on at least one of these paths?

$\textbf{(A)}\ 161 \qquad \textbf{(B)}\ 185 \qquad \textbf{(C)}\ 195 \qquad \textbf{(D)}\ 227 \qquad \textbf{(E)}\ 255$

Solution

We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20$, then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$, and $|x + 3| + |y - 2|$ is the length of the shortest path from $(x,y)$ to $(-3, 2)$


If $-3\le x \le 3$, then $-7\le y \le 7$ satisfy the property. there are $15 \times 7 = 105$ lattice points here.

else let $3< x \le 8$ (and for $-8 \le x < -3$ it is symmetrical) $-7 + (x - 3)\le y \le 7 - (x - 3)$,

$-10 + x\le y \le 10 - x$

So for $x = 4$, there are $13$ lattice points,

for $x = 5$, there are $11$ lattice points,

etc.

For $x = 8$, there are $5$ lattice points.

Hence, there are a total of $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}$ lattice points. $\square$

One may also obtain the result by using Pick's Theorem.

See also

2011 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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