Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem 11)
(→‎Problem 11)
Line 7: Line 7:
 
==Solution 1==
 
==Solution 1==
  
−
The <math>3\times4</math> rectangle allows for 4 possible places to put the S piece, with each possible placement having an inverted version.
+
The <math>3\times4</math> rectangle allows for <math>7</math> possible places to put the S piece, with each possible placement having an inverted version.
 +
One of the cases looks like this:
 
<asy>
 
<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 <math>3</math> 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>
  
−
path x = (0,0)--(0,2)--(1,2)--(1,3)--(2,3)--(2,1)--(1,1)--(1,0)--cycle;
+
<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));
 
add(grid(4,3));
−
fill(x; gray(1));
 
 
</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

Revision as of 12:41, 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?

$\textbf{(A)}I$ 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_8_Problems/Problem_11&oldid=241196"