Difference between revisions of "2011 AMC 12B Problems/Problem 23"
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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 21:50, 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. 
