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

Revision as of 22:21, 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 $x = 4$ , 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?).

@@ Line 15: / Line 15: @@
 Now we test each value for x satisfying <math>3< x \le 8</math> and double the result because of symmetry.
-For <math>x = 4</math>, there are <math>13</math> lattice points,
+For <math>x = 4</math>, the possibles values of y are such that <math>|2y| \le 12</math> for a total of <math>13</math> lattice points,
-for <math>x = 5</math>, there are <math>11</math> lattice points,
+for <math>x = 5</math>, the possibles values of y are such that <math>|2y| \le 10</math> for a total of <math>11</math> lattice points,
-for <math>x = 6</math>, there are <math>9</math> lattice points,
+for <math>x = 6</math>, the possibles values of y are such that <math>|2y| \le 8</math> for a total of <math>9</math> lattice points,
-for <math>x = 7</math>, there are <math>7</math> lattice points,
+for <math>x = 7</math>, the possibles values of y are such that <math>|2y| \le 6</math> for a total of <math>7</math> lattice points,
-For <math>x = 8</math>, there are <math>5</math> lattice points.
+for <math>x = 8</math>, the possibles values of y are such that <math>|2y| \le 4</math> for a total of <math>5</math> lattice points,
 <br />

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

Revision as of 22:21, 19 January 2020

Problem

Solution

See also