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Difference between revisions of "2023 AMC 12B Problems/Problem 5"

(→‎Solution 1)
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draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0));
 
draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0));
 
label("7",(0.5,0.5));
 
label("7",(0.5,0.5));
−
draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1), red);
 
 
label("8",(1.5,0.5));
 
label("8",(1.5,0.5));
 
draw((0,2)--(1,2)--(1,3)--(0,3)--(0,2));
 
draw((0,2)--(1,2)--(1,3)--(0,3)--(0,2));
 
label("9",(2.5,0.5));
 
label("9",(2.5,0.5));
−
draw((1,0)--(2,0)--(2,1)--(1,1)--(1,0),red);
 
 
label("4",(0.5,1.5));
 
label("4",(0.5,1.5));
 
draw((1,1)--(2,1)--(2,2)--(1,2)--(1,1));
 
draw((1,1)--(2,1)--(2,2)--(1,2)--(1,1));
 
label("5",(1.5,1.5));
 
label("5",(1.5,1.5));
−
draw((1,2)--(2,2)--(2,3)--(1,3)--(1,2),red);
 
 
label("6",(2.5,1.5));
 
label("6",(2.5,1.5));
 
draw((2,0)--(3,0)--(3,1)--(2,1)--(2,0));
 
draw((2,0)--(3,0)--(3,1)--(2,1)--(2,0));
 
label("1",(0.5,2.5));
 
label("1",(0.5,2.5));
−
draw((2,1)--(3,1)--(3,2)--(2,2)--(2,1),red);
 
 
label("2",(1.5,2.5));
 
label("2",(1.5,2.5));
 
draw((2,2)--(3,2)--(3,3)--(2,3)--(2,2));
 
draw((2,2)--(3,2)--(3,3)--(2,3)--(2,2));
 
label("3",(2.5,2.5));
 
label("3",(2.5,2.5));
 +
 +
draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1), red);
 +
draw((1,0)--(2,0)--(2,1)--(1,1)--(1,0),red);
 +
draw((1,2)--(2,2)--(2,3)--(1,3)--(1,2),red);
 +
draw((2,1)--(3,1)--(3,2)--(2,2)--(2,1),red);
 +
 
</asy>
 
</asy>
  

Revision as of 19:48, 15 November 2023

The following problem is from both the 2023 AMC 10B #10 and 2023 AMC 12B #5, so both problems redirect to this page.

Problem

You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$

Solution 1

[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); label("7",(0.5,0.5)); draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1)); label("8",(1.5,0.5)); draw((0,2)--(1,2)--(1,3)--(0,3)--(0,2)); label("9",(2.5,0.5)); draw((1,0)--(2,0)--(2,1)--(1,1)--(1,0)); label("4",(0.5,1.5)); draw((1,1)--(2,1)--(2,2)--(1,2)--(1,1)); label("5",(1.5,1.5)); draw((1,2)--(2,2)--(2,3)--(1,3)--(1,2)); label("6",(2.5,1.5)); draw((2,0)--(3,0)--(3,1)--(2,1)--(2,0)); label("1",(0.5,2.5)); draw((2,1)--(3,1)--(3,2)--(2,2)--(2,1)); label("2",(1.5,2.5)); draw((2,2)--(3,2)--(3,3)--(2,3)--(2,2)); label("3",(2.5,2.5)); [/asy]


First, we notice $12 \text{ } 2\times1$ rectangles in a $3\times3$ grid.

Next, if we choose one of the corners (1,3,7,9), and the corner is not covered by a $2 \times 1$ rectangle, we can eliminate a maximum of 2 rectangles.

If we choose one of the side squares (2,4,6,8), we can eliminate a maximum of $3$ rectangles.

Finally, if we choose the center square (5), we can eliminate a maximum of $4$ rectangles, but doing so means that if we choose a side square, we only eliminate 2 rectangles.

Now, we want to sum to 12 only by using $2,3,4$. If we have a $4,$ (i.e. we choose the middle square,) the remaining numbers can only be $2.$ So, we find that $4+2\times4=12,$ meaning we need $1+4=5$ turns to guarantee that a square is covered.

If we don't use the middle square, we find that we only need $12 \div 3 = 4$ turns. This means we choose all four sides: [asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); label("7",(0.5,0.5)); label("8",(1.5,0.5)); draw((0,2)--(1,2)--(1,3)--(0,3)--(0,2)); label("9",(2.5,0.5)); label("4",(0.5,1.5)); draw((1,1)--(2,1)--(2,2)--(1,2)--(1,1)); label("5",(1.5,1.5)); label("6",(2.5,1.5)); draw((2,0)--(3,0)--(3,1)--(2,1)--(2,0)); label("1",(0.5,2.5)); label("2",(1.5,2.5)); draw((2,2)--(3,2)--(3,3)--(2,3)--(2,2)); label("3",(2.5,2.5)); draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1), red); draw((1,0)--(2,0)--(2,1)--(1,1)--(1,0),red); draw((1,2)--(2,2)--(2,3)--(1,3)--(1,2),red); draw((2,1)--(3,1)--(3,2)--(2,2)--(2,1),red); [/asy]

The answer is $\boxed{(C) 4}$

Solution 2

First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ guesses, and one with $4$. Since the problem is asking for the minimum number, the answer is $\boxed{\text{(C) } 4}$.

~yourmomisalosinggame (a.k.a. Aaron)

Solution 3

Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\boxed{\text{(C) } 4}$

~arrowskyknight22

See Also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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
2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

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