Difference between revisions of "2011 AMC 12B Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | + | 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>. | |
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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. | ||
− | + | 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. | |
− | + | Now we test each value for x satisfying <math>3< x \le 8</math> and double the result because of symmetry. | |
− | </math> | + | 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>, the possibles values of y are such that <math>|2y| \le 10</math> for a total of <math>11</math> lattice points, | |
− | for < | + | 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>, the possibles values of y are such that <math>|2y| \le 6</math> for a total of <math>7</math> lattice points, | |
− | for < | + | 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 /> | <br /> | ||
− | Hence, there are a total of < | + | Hence, there are a total of <math>105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}</math> lattice points. |
+ | |||
+ | One may also obtain the result by using Pick's Theorem(how?). | ||
== 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}} |
Latest revision as of 23:21, 19 January 2020
Problem
A bug travels in the coordinate plane, moving only along the lines that are parallel to the -axis or -axis. Let and . Consider all possible paths of the bug from to of length at most . How many points with integer coordinates lie on at least one of these paths?
Solution
We declare a point to make up for the extra steps that the bug has to move. If the point satisfies the property that , then it is in the desirable range because is the length of the shortest path from to and is the length of the shortest path from to .
If , then satisfy the property. there are lattice points here.
else let (and for because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from to added to the shortest distance from to is . Since the minimum value for the difference between the y-coordinates is at , we get or . Thus, the upper and lower bounds for are and , respectively.
Now we test each value for x satisfying and double the result because of symmetry.
For , the possibles values of y are such that for a total of lattice points,
for , the possibles values of y are such that for a total of lattice points,
for , the possibles values of y are such that for a total of lattice points,
for , the possibles values of y are such that for a total of lattice points,
for , the possibles values of y are such that for a total of lattice points,
Hence, there are a total of lattice points.
One may also obtain the result by using Pick's Theorem(how?).
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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