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Difference between revisions of "2025 AMC 8 Problems/Problem 11"

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A <math>\textit{tetromino}</math> consists of four squares connected along their edges. There are five possible tetromino shapes, <math>I</math>, <math>O</math>, <math>L</math>, <math>T</math>, and <math>S</math>, shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a <math>3\times4</math> rectangle. At least one of the tiles is an <math>S</math> tile. What are the other two tiles?
 
A <math>\textit{tetromino}</math> consists of four squares connected along their edges. There are five possible tetromino shapes, <math>I</math>, <math>O</math>, <math>L</math>, <math>T</math>, and <math>S</math>, shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a <math>3\times4</math> rectangle. At least one of the tiles is an <math>S</math> tile. What are the other two tiles?
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<asy>
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unitsize(12);
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add(grid(1,4));
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label("I", (0.5,-1));
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add(shift((5,0)) * grid(2,2));
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label("O", (6,-1));
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add(shift((11,0)) * grid(1,3));
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add(shift((11,0)) * grid(2,1));
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label("L", (12,-1));
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add(shift((18,0)) * grid(1,1));
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add(shift((17,1)) * grid(3,1));
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label("T", (18.5,-1));
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add(shift((25,1)) * grid(2,1));
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add(shift((24,0)) * grid(2,1));
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label("S", (25.5,-1));
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add(shift((12,-6)) * grid(4,3));
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</asy>
  
 
<math>\textbf{(A)}I</math> and <math>L\qquad \textbf{(B)} I</math> and <math>T\qquad \textbf{(C)} L</math> and <math>L\qquad \textbf{(D)}L</math> and <math>S\qquad \textbf{(E)}O</math> and <math>T</math>
 
<math>\textbf{(A)}I</math> and <math>L\qquad \textbf{(B)} I</math> and <math>T\qquad \textbf{(C)} L</math> and <math>L\qquad \textbf{(D)}L</math> and <math>S\qquad \textbf{(E)}O</math> and <math>T</math>
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</asy>
 
</asy>
 
For each of the inverted cases, the L pieces can be inverted along with the S piece. Therefore, because the only cases that fill the rectangle after the S is placed are the ones that use two L pieces, the answer must be <math>\boxed{\textbf{(C)}~L \ and \ L}</math>. ~bubby617
 
For each of the inverted cases, the L pieces can be inverted along with the S piece. Therefore, because the only cases that fill the rectangle after the S is placed are the ones that use two L pieces, the answer must be <math>\boxed{\textbf{(C)}~L \ and \ L}</math>. ~bubby617
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 20:15, 30 January 2025

Problem 11

A $\textit{tetromino}$ consists of four squares connected along their edges. There are five possible tetromino shapes, $I$, $O$, $L$, $T$, and $S$, shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a $3\times4$ rectangle. At least one of the tiles is an $S$ tile. What are the other two tiles?

[asy] unitsize(12); add(grid(1,4)); label("I", (0.5,-1)); add(shift((5,0)) * grid(2,2)); label("O", (6,-1)); add(shift((11,0)) * grid(1,3)); add(shift((11,0)) * grid(2,1)); label("L", (12,-1)); add(shift((18,0)) * grid(1,1)); add(shift((17,1)) * grid(3,1)); label("T", (18.5,-1)); add(shift((25,1)) * grid(2,1)); add(shift((24,0)) * grid(2,1)); label("S", (25.5,-1)); add(shift((12,-6)) * grid(4,3)); [/asy]

$\textbf{(A)}I$ and $L\qquad \textbf{(B)} I$ and $T\qquad \textbf{(C)} L$ and $L\qquad \textbf{(D)}L$ and $S\qquad \textbf{(E)}O$ and $T$

Solution 1

The $3\times4$ rectangle allows for $7$ possible places to put the S piece, with each possible placement having an inverted version. One of the cases looks like this: [asy] path x = (0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,1)--(1,1)--(1,0)--cycle; fill(x, rgb(0,30,0)); add(grid(4,3)); [/asy] Now, as you can see, there's a hole in the top left corner of the board, which would be impossible to fill using the tetrominos. There are $3$ cases in which one of these holes isn't created; the S lies flat in the bottom left corner, it lies flat in the top right corner, or it stands upright in the center. In all three cases, two L's can be used to fill the remaining empty spaces in the rectangle, shown below. [asy] path z1 = (2,0)--(3,0)--(3,2)--(2,2)--(2,3)--(1,3)--(1,1)--(2,1)--cycle; path z2 = (0,0)--(2,0)--(2,1)--(1,1)--(1,3)--(0,3)--cycle; path z3 = (2,3)--(4,3)--(4,0)--(3,0)--(3,2)--(2,2)--cycle; fill(z1, rgb(0,30,0)); fill(z2, rgb(127,80,0)); fill(z3, rgb(127,100,0)); add(grid(4,3)); [/asy]

[asy] path y1 = (0,0)--(2,0)--(2,1)--(3,1)--(3,2)--(1,2)--(1,1)--(0,1)--cycle; path y2 = (0,1)--(1,1)--(1,2)--(3,2)--(3,3)--(0,3)--cycle; path y3 = (2,0)--(4,0)--(4,3)--(3,3)--(3,1)--(2,1)--cycle; fill(y1, rgb(0,30,0)); fill(y2, rgb(127,80,0)); fill(y3, rgb(127,100,0)); add(grid(4,3)); [/asy]

[asy] path w1 = (1,1)--(3,1)--(3,2)--(4,2)--(4,3)--(2,3)--(2,2)--(1,2)--cycle; path w2 = (0,0)--(0,3)--(2,3)--(2,2)--(1,2)--(1,0)--cycle; path w3 = (1,0)--(4,0)--(4,2)--(3,2)--(3,1)--(1,1)--cycle; fill(w1, rgb(0,30,0)); fill(w2, rgb(127,80,0)); fill(w3, rgb(127,100,0)); add(grid(4,3)); [/asy] For each of the inverted cases, the L pieces can be inverted along with the S piece. Therefore, because the only cases that fill the rectangle after the S is placed are the ones that use two L pieces, the answer must be $\boxed{\textbf{(C)}~L \ and \ L}$. ~bubby617

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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