Difference between revisions of "2011 AMC 12B Problems/Problem 23"
(→Solution) |
m (→Solution) |
||
Line 26: | Line 26: | ||
Hence, there are a total of <math>105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}</math> lattice points. <math>\square</math> | Hence, there are a total of <math>105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}</math> lattice points. <math>\square</math> | ||
− | One may also obtain the result by using Pick's Theorem. | + | 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}} | {{MAA Notice}} |
Revision as of 22:50, 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 it is symmetrical) ,
So for , there are lattice points,
for , there are lattice points,
etc.
For , there are 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.