Difference between revisions of "2022 AMC 10B Problems/Problem 19"

Revision as of 05:05, 26 November 2022

The following problem is from both the 2022 AMC 10B #19 and 2022 AMC 12B #18, so both problems redirect to this page.

Problem

Each square in a $5 \times 5$ 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. $[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]$ Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ 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.) $[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]$ $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$

Solution

There are two cases for the initial configuration:

The center square is filled.

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,gray+opacity(0.5),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]$ In this case, there are $2$ possible initial configurations. All rotations and reflections are considered.

The center square is empty.

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,gray+opacity(0.5),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]$ In this case, there are $4+4+4+8=20$ possible initial configurations. All rotations and reflections are considered.

Together, the answer is $2+20=\boxed{\textbf{(C)}\ 22}.$

~mathboy100 ~MRENTHUSIASM

Video Solution by OmegaLearn (Using Logic and Casework)

https://youtu.be/UYTiP3u5qE4

~ pi_is_3.14

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

Page

Toolbox

Search

Difference between revisions of "2022 AMC 10B Problems/Problem 19"

Revision as of 05:05, 26 November 2022

Contents

Problem

Solution

Video Solution by OmegaLearn (Using Logic and Casework)

See Also

@@ Line 86: / Line 86: @@
 	filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible);
 }
+ds((1,3));
+ds((2,2));
+ds((3,1));
 for (int i = 0; i <= 5; ++i) {
@@ Line 92: / Line 96: @@
      }
-ds((1,3));
-ds((2,2));
-ds((3,1));
 label("$2$ Configurations", (2.5,-1));
 </asy>
@@ Line 107: / Line 108: @@
 	filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible);
 }
-for (int i = 0; i <= 5; ++i) {
-	draw((0,i)--(5,i));
-    draw((i,0)--(i,5));
-    }
 ds((1,3));
@@ Line 118: / Line 114: @@
 for (int i = 0; i <= 5; ++i) {
-    draw((9,i)--(14,i));
+	draw((0,i)--(5,i));
-     draw((i+9,0)--(i+9,5));
+     draw((i,0)--(i,5));
      }
@@ Line 127: / Line 123: @@
 for (int i = 0; i <= 5; ++i) {
-     draw((18,i)--(23,i));
+     draw((9,i)--(14,i));
-     draw((i+18,0)--(i+18,5));
+     draw((i+9,0)--(i+9,5));
      }
@@ Line 136: / Line 132: @@
 for (int i = 0; i <= 5; ++i) {
-     draw((27,i)--(32,i));
+     draw((18,i)--(23,i));
-     draw((i+27,0)--(i+27,5));
+     draw((i+18,0)--(i+18,5));
      }
@@ Line 143: / Line 139: @@
 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));