Difference between revisions of "2011 AMC 12B Problems/Problem 11"
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| − | Thus, the answer is <cmath> = \boxed{3\ \(\textbf{( | + | Thus, the answer is <cmath> = \boxed{3\ \(\textbf{(B)}} </cmath> |
{{AMC12 box|year=2011|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2011|ab=B|num-b=10|num-a=12}} | ||
Revision as of 15:31, 5 June 2011
Problem
A frog located at 
, with both 
and 
integers, makes successive jumps of length 
and always lands on points with integer coordinates. Suppose that the frog starts at 
and ends at 
. What is the smallest possible number of jumps the frog makes?
Solution
Since the frog always jumps in length and lands on a lattice point, the sum of its coordinates must change either by (by jumping parallel to the x- or y-axis), or by 
or 
(by making a move similar to a knight on a chessboard, with two units along one axis and one unit along the other.)
Because either , , or is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it is impossible for the frog to go from to in an even number of moves. Therefore, the frog cannot reach in two moves.
However, a path is possible in 3 moves: from to 
to 
to .
Thus, the answer is
\[= \boxed{3\ \(\textbf{(B)}}\] (Error compiling LaTeX. Unknown error_msg)
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 | |
