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"

(→‎Solution)
(Just a suggestion (I don't know how to apply it))
Line 29: Line 29:
  
 
One may also obtain the result by using Pick's Theorem(how?).
 
One may also obtain the result by using Pick's Theorem(how?).
 +
 +
<math>i = a - \frac{b}{2} - 1</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=22|num-a=24|ab=B}}
 
{{AMC12 box|year=2011|num-b=22|num-a=24|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:36, 1 June 2022

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 $-8 \le x < -3$ because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from $(-3, 2)$ to $(x, y)$ added to the shortest distance from $(x, y)$ to $(3, -2)$ is $|x - 3| + |y + 2| + |x + 3| + |y - 2|$. Since the minimum value for the difference between the y-coordinates is at $y = 0$, we get $2x + 4 = 16$ or $-2x + 4 = 16$. Thus, the upper and lower bounds for $x$ are $8$ and $-8$, respectively.

Now we test each value for x satisfying $3< x \le 8$ and double the result because of symmetry.

For $x = 4$, the possibles values of y are such that $|2y| \le 12$ for a total of $13$ lattice points,

for $x = 5$, the possibles values of y are such that $|2y| \le 10$ for a total of $11$ lattice points,

for $x = 6$, the possibles values of y are such that $|2y| \le 8$ for a total of $9$ lattice points,

for $x = 7$, the possibles values of y are such that $|2y| \le 6$ for a total of $7$ lattice points,

for $x = 8$, the possibles values of y are such that $|2y| \le 4$ for a total of $5$ lattice points,

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

One may also obtain the result by using Pick's Theorem(how?).

$i = a - \frac{b}{2} - 1$

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=174221"