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Difference between revisions of "2023 AMC 12B Problems/Problem 5"
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==Solution 1==
 
==Solution 1==
−
<asy>
+
Notice that the <math>3\times3</math> square grid has a total of <math>12</math> possible <math>2\times1</math> rectangles.
−
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 <math>12 \text{ } 2\times1</math> rectangles in a <math>3\times3</math> grid.
 
−
 
 
−
Next, if we choose one of the corners (1,3,7,9), and the corner is not covered by a <math>2 \times 1</math> 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 <math>3</math> rectangles.  
+
Suppose you choose the middle square for one of your turns. The middle square is covered by <math>4</math> rectangles, and each of the remaining <math>8</math> squares is covered by a maximum of <math>2</math> uncounted rectangles. This means that the number of turns is at least <math>1+\frac{12-4}{2}=1+4=5</math>.
  
−
Finally, if we choose the center square (5), we can eliminate a maximum of <math>4</math> rectangles, but doing so means that if we choose a side square, we only eliminate 2 rectangles.
+
Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is <math>12\div 3=4.</math> To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a <math>T</math> shape and they do not cover any other side square. Hence, our answer is <math>4.</math>
  
−
Now, we want to sum to 12 only by using <math>2,3,4</math>. If we have a <math>4,</math> (i.e. we choose the middle square,) the remaining numbers can only be <math>2.</math> So, we find that <math>4+2\times4=12,</math> meaning we need <math>1+4=5</math> turns to guarantee that a square is covered.
 
  
−
If we don't use the middle square, we find that we only need <math>12 \div 3 = 4</math> turns. This means we choose all four sides:
 
 
<asy>
 
<asy>
−
draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0));
+
draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0));
−
label("7",(0.5,0.5));
+
draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1));
−
label("8",(1.5,0.5));
+
draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5));
−
draw((0,2)--(1,2)--(1,3)--(0,3)--(0,2));
+
draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0));
−
label("9",(2.5,0.5));
+
draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1));
−
label("4",(0.5,1.5));
+
draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5));
−
draw((1,1)--(2,1)--(2,2)--(1,2)--(1,1));
+
draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0));
−
label("5",(1.5,1.5));
+
draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1));
−
label("6",(2.5,1.5));
+
draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5));
−
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);
 
  
 +
filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1));
 +
filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1));
 +
filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1));
 +
filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1));
 
</asy>
 
</asy>
−
 
−
The answer is <math>\boxed{(C) 4}</math>
 
  
 
==Solution 2==
 
==Solution 2==
Line 69: Line 35:
 
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 <math>5</math> guesses, and one with <math>4</math>. Since the problem is asking for the minimum number, the answer is <math>\boxed{\text{(C) }  4}</math>.
 
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 <math>5</math> guesses, and one with <math>4</math>. Since the problem is asking for the minimum number, the answer is <math>\boxed{\text{(C) }  4}</math>.
  
−
~yourmomisalosinggame (a.k.a. Aaron)
+
~Stead (a.k.a. Aaron)
  
 
==Solution 3==
 
==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 :
+
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 can't be diagonal, we can ensure that we find it in 4 moves. So our answer is :
 
<math>\boxed{\text{(C) }  4}</math>
 
<math>\boxed{\text{(C) }  4}</math>
  
 
~arrowskyknight22
 
~arrowskyknight22
  
−
==Video Solution 1 by OmegaLearn==
+
==Solution 4 (checkerboard)==
 +
The <math>3 \times 3</math> grid can be colored like a checkerboard with alternating black and white squares.
 +
Let the top left square be white, and the rest of the squares alternate colors.
 +
 
 +
Each <math>2 \times 1</math> rectangle always covers <math>1</math> white square and <math>1</math> black square.
 +
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares (<math>5</math> turns) or each of the black squares (<math>4</math> turns).
 +
 
 +
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the <math>\boxed{4}</math> squares are of the <math>2 \times 1</math> rectangle.
 +
 
 +
~ CherryBerry
 +
 
 +
(Minor edits by NSAoPS)
 +
 
 +
==Solution 5 (Logic)==
 +
We realize that every <math>2 \times 1</math> rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is <math>\boxed{\text{(C) }  4.}</math>.
 +
~darrenn.cp
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/EuLkw8HFdk4?si=RbKZpmAv9kfDLuON&t=2106
 +
 
 +
~Math-X
 +
 
 +
==Video Solution 1 by MegaMath==
 +
https://www.youtube.com/watch?v=rMZrqGD0MKg&t=3s
 +
~megahertz13
 +
 
 +
==Video Solution 2 by OmegaLearn==
 
https://youtu.be/7rO2-mtQCvM
 
https://youtu.be/7rO2-mtQCvM
  
 +
==Video Solution 3==
 +
https://youtu.be/EvA2Nlb7gi4?si=fVLG8gMTIC5XkEwP
 +
 +
==Video Solution==
 +
 +
https://youtu.be/piiG6JAgxAs
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
  
 
==See Also==
 
==See Also==

Latest revision as of 18:46, 28 October 2024

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

Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles.

Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$.

Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$


[asy] draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)); draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)); draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)); draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0)); draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1)); draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)); draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0)); draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1)); draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5)); filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1)); filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1)); filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1)); filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1)); [/asy]

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 $\boxed{\text{(C) } 4}$.

~Stead (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 can't be diagonal, we can ensure that we find it in 4 moves. So our answer is : $\boxed{\text{(C) } 4}$

~arrowskyknight22

Solution 4 (checkerboard)

The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares. Let the top left square be white, and the rest of the squares alternate colors.

Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square. You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ($5$ turns) or each of the black squares ($4$ turns).

Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{4}$ squares are of the $2 \times 1$ rectangle.

~ CherryBerry

(Minor edits by NSAoPS)

Solution 5 (Logic)

We realize that every $2 \times 1$ rectangle must contain an edge and no more than one edge. There are a total of four edges so the answer is $\boxed{\text{(C) } 4.}$. ~darrenn.cp

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=RbKZpmAv9kfDLuON&t=2106

~Math-X

Video Solution 1 by MegaMath

https://www.youtube.com/watch?v=rMZrqGD0MKg&t=3s ~megahertz13

Video Solution 2 by OmegaLearn

https://youtu.be/7rO2-mtQCvM

Video Solution 3

https://youtu.be/EvA2Nlb7gi4?si=fVLG8gMTIC5XkEwP

Video Solution

https://youtu.be/piiG6JAgxAs

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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