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

m (Solution)
(Solution)
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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> because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from <math>(-3, 2)</math> to <math>(x, y)</math> added to the shortest distance from <math>(x, y)</math> to <math>(3, -2)</math> is <math>|x - 3| + |y + 2| + |x + 3| + |y - 2|</math>. Since the minimum value for the difference between the y-coordinates is at <math>y = 0</math>, we get <math>2x + 4 = 16</math> or <math>-2x + 4 = 16</math>. Thus, the upper and lower bounds for <math>x</math> are <math>8</math> and <math>-8</math>, respectively.
  
<math>-10 + x\le y \le 10 - x</math>
+
Now we test each value for x satisfying <math>3< x \le 8</math> and double the result because of symmetry.
  
So for <math>x = 4</math>, there are <math>13</math> lattice points,
+
For <math>x = 4</math>, there are <math>13</math> lattice points,
  
 
for <math>x = 5</math>, there are <math>11</math> lattice points,
 
for <math>x = 5</math>, there are <math>11</math> lattice points,
  
etc.
+
for <math>x = 6</math>, there are <math>9</math> lattice points,
 +
 
 +
for <math>x = 7</math>, there are <math>7</math> lattice points,
  
 
For <math>x = 8</math>, there are <math>5</math> lattice points.
 
For <math>x = 8</math>, there are <math>5</math> lattice points.

Revision as of 23:10, 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 $-8 \le x < -3$ 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 $x = 4$, there are $13$ lattice points,

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

for $x = 6$, there are $9$ lattice points,

for $x = 7$, there are $7$ lattice points,

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(how?).

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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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