Difference between revisions of "2023 AMC 8 Problems/Problem 23"

Revision as of 10:54, 17 February 2023

Problem

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right. $[asy] size(5.663333333cm); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill((6,.33)--(7,.33)--(7,1.33)--cycle,mediumgray); draw((6,.33)--(7,.33)--(7,1.33)--(6,1.33)--cycle,gray); fill((6,1.67)--(7,2.67)--(6,2.67)--cycle,mediumgray); draw((6,1.67)--(7,1.67)--(7,2.67)--(6,2.67)--cycle,gray); fill((7.33,.33)--(8.33,.33)--(7.33,1.33)--cycle,mediumgray); draw((7.33,.33)--(8.33,.33)--(8.33,1.33)--(7.33,1.33)--cycle,gray); fill((8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,mediumgray); draw((7.33,1.67)--(8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,gray); [/asy]$ What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling. $[asy] size(2cm); fill((1,0)--(0,1)--(0,2)--(1,1)--cycle,mediumgray); fill((2,0)--(3,1)--(2,2)--(1,1)--cycle,mediumgray); fill((1,2)--(1,3)--(0,3)--cycle,mediumgray); fill((1,2)--(2,2)--(2,3)--cycle,mediumgray); fill((3,2)--(3,3)--(2,3)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); [/asy]$

$\textbf{(A) } \frac{1}{1024} \qquad \textbf{(B) } \frac{1}{256} \qquad \textbf{(C) } \frac{1}{64} \qquad \textbf{(D) } \frac{1}{16} \qquad \textbf{(E) } \frac{1}{4}$

Solution 1

There are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids, as shown below: $[asy] size(375); fill((1,1)--(2,2)--(1,3)--(0,2)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill(shift(7,0)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(6,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(6,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(6,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(12,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(12,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(12,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(12,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(19,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(18,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(18,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(18,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); [/asy]$ There are $4^5$ ways to decide the $5$ white squares for each case, and the cases do not have any overlap.

So, the requested probability is $\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.$ -apex304, TaeKim, MRENTHUSIASM

Solution 2

Note that the middle tile can be any of the four tiles. The gray part of the middle tile points towards one of the corners, and for the gray diamond to appear the three adjacent tiles must all be perfect. Thus, the solution is $\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\textbf{(C) } \frac{1}{64}}$ .

~aayr

Solution 3 (Linearity of Expectation)

Let $S_1, S_2, S_3$ , and $S_4$ denote the $4$ smaller $2 \times 2$ squares within the $3 \times 3$ square in some order. For each $S_i$ , let $X_i = 1$ if it contains a large gray diamond tiling and $X_i = 0$ otherwise. This means that $\mathbb{E}[X_i]$ is the probability that square $S_i$ has a large gray diamond, so $\mathbb{E}[X_1 + X_2 + X_3 + X_4]$ is our desired probability. However, since there is only one possible way to arrange the squares within every $2 \times 2$ square to form such a tiling, we have $\mathbb{E}[X_i] = (\tfrac{1}{4})^2 = \tfrac{1}{256}$ for all $i$ (as each of the smallest tiles has $4$ possible arrangements), and from the linearity of expectation we get $\mathbb{E}[X_1 + X_2 + X_3 + X_4] = \mathbb{E}[X_1] + \mathbb{E}[X_2] + \mathbb{E}[X_3] + \mathbb{E}[X_4] = \frac{1}{256} + \frac{1}{256} + \frac{1}{256} + \frac{1}{256} = \boxed{\textbf{(C) } \frac{1}{64}}.$ ~eibc

Remark 1: This method might be too advanced for the AMC 8, and is probably unnecessary (refer to the other solutions for simpler techniques).

Remark 2: Note that Probability and Expected Value are equivalent in this problem since there will never be two diamonds on one tiling. i.e. $X_1 + X_2 + X_3 + X_4 \le 1$ .

~numerophile

Solution 4

We notice that there is only one arrangement of 4 tiles that works. Therefore, they must be perfectly selected. We have a

\[\frac{1⁵}{4⁵}\] (Error compiling LaTeX. Unknown error_msg)

chance of getting it right, and 4 different places to place the 2 by 2 square, so

\[\frac{1}{4⁵} \cdot 4 = \frac{1}{4⁴} = \boxed{\textbf{(C) } \frac{1}{64}}\] (Error compiling LaTeX. Unknown error_msg)

Video Solution 1 by OmegaLearn (Using Cool Probability Technique)

https://youtu.be/2t_Za0Y2IqY

Animated Video Solution

https://youtu.be/f4ffQEG0yUw

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=2405

Video Solution by Interstigation

https://youtu.be/1bA7fD7Lg54?t=2338

@@ Line 79: / Line 79: @@
 ~numerophile
-==Solution 4
+==Solution 4==
-We notice that there is only one arrangement of 4 tiles that works. Therefore, they must be perfectly selected. We have a <cmath>\frac14^5</cmath>
+We notice that there is only one arrangement of 4 tiles that works. Therefore, they must be perfectly selected. We have a <cmath>\frac{1⁵}{4⁵}</cmath> chance of getting it right, and 4 different places to place the 2 by 2 square, so <cmath>\frac{1}{4⁵} \cdot 4 = \frac{1}{4⁴} = \boxed{\textbf{(C) } \frac{1}{64}}</cmath>
 ==Video Solution 1 by OmegaLearn (Using Cool Probability Technique)==

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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