Difference between revisions of "2022 AMC 10B Problems/Problem 19"
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void ds(pair x) { | void ds(pair x) { | ||
| − | filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle, | + | filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); |
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void ds(pair x) { | void ds(pair x) { | ||
| − | filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle, | + | filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); |
} | } | ||
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~MathProblemSolvingSkills.com | ~MathProblemSolvingSkills.com | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/JVDlHCSPF6k | ||
| + | |||
| + | ~Hayabusa1 | ||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/gsaD0wQPVgY | ||
| + | |||
| + | ~Interstigation | ||
== See Also == | == See Also == | ||
Latest revision as of 02:22, 7 July 2023
- The following problem is from both the 2022 AMC 10B #19 and 2022 AMC 12B #18, so both problems redirect to this page.
Contents
Problem
Each square in a 
grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
- Any filled square with two or three filled neighbors remains filled.
- Any empty square with exactly three filled neighbors becomes a filled square.
- All other squares remain empty or become empty.
A sample transformation is shown in the figure below.
![[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]](http://latex.artofproblemsolving.com/0/0/3/00333423a7ddd483233c145023df02bdb469a3b9.png)
Suppose the grid has a border of empty squares surrounding a 
subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)
![[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]](http://latex.artofproblemsolving.com/2/5/3/253c2691ed603d5d600e2b1becee941d5268c900.png)
Solution
There are two cases for the initial configuration:
- The center square is filled.
- The center square is empty.
Exactly two of the eight adjacent neighboring squares of the center are filled. Clearly, the only possibility is that the squares along one diagonal are filled, as shown below:
![[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((2,2)); ds((3,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("$2$ Configurations", (2.5,-1)); [/asy]](http://latex.artofproblemsolving.com/8/d/9/8d9ca05cb188cd4c1903e7c6ced4882797e1db84.png)
In this case, there are 
possible initial configurations. All rotations and reflections are considered.
Exactly three of the eight adjacent neighboring squares of the center are filled. The possibilities are shown below:
![[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((3,3)); ds((1,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } ds((10,3)); ds((12,3)); ds((11,1)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } ds((19,3)); ds((20,1)); ds((21,2)); for (int i = 0; i <= 5; ++i) { draw((18,i)--(23,i)); draw((i+18,0)--(i+18,5)); } ds((28,3)); ds((29,1)); ds((30,1)); for (int i = 0; i <= 5; ++i) { draw((27,i)--(32,i)); draw((i+27,0)--(i+27,5)); } label("$4$ Configurations", (2.5,-1)); label("$4$ Configurations", (11.5,-1)); label("$4$ Configurations", (20.5,-1)); label("$8$ Configurations", (29.5,-1)); [/asy]](http://latex.artofproblemsolving.com/9/a/f/9af0bc0ac9bf851c3078feeda3696d8c643a3b6d.png)
In this case, there are 
possible initial configurations. All rotations and reflections are considered.
Together, the answer is 
~mathboy100 ~MRENTHUSIASM
Video Solution by OmegaLearn (Using Logic and Casework)
~ pi_is_3.14
Video Solution
~MathProblemSolvingSkills.com
Video Solution
~Hayabusa1
Video Solution by Interstigation
~Interstigation
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 10 Problems and Solutions | ||
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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. 
