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 "2011 AMC 12B Problems/Problem 23"
Line 13: Line 13:
 
If <math>-3\le x \le 3</math>, then <math>-7\le y \le 7</math> satisfy the property. there are <math>15 \times 7 = 105</math> lattice points here.
 
If <math>-3\le x \le 3</math>, then <math>-7\le y \le 7</math> satisfy the property. there are <math>15 \times 7 = 105</math> lattice points here.
  
−
else let <math>3< x \le 8</math> (and for <math>-8 \le x < -3</math> it is symmetrical, <math>-7 + (x - 3)\le y \le 7 - (x - 3)</math>,
+
else let <math>3< x \le 8</math> (and for <math>-8 \le x < -3</math> it is symmetrical) <math>-7 + (x - 3)\le y \le 7 - (x - 3)</math>,
  
 
<math>-10 + x\le y \le 10 - x</math>
 
<math>-10 + x\le y \le 10 - x</math>

Revision as of 14:59, 6 January 2018

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

Answer: (C)

If a point $(x, y)$ satisfy 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 shortest path from $(x,y)$ to $B$, and $|x + 3| + |y - 2|$ is the shortest path from $(x,y)$ to $A$


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 $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{195}$ lattice points.

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

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=2011_AMC_12B_Problems/Problem_23&oldid=89484"