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

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<math>\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}</math>
 
<math>\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}</math>
  
 +
==Solution 1==
 +
There are <math>4</math> cases that the tiling will contain a large gray diamond in one of the smaller <math>2 \times 2</math> 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 <math>4^5</math> ways to decide the <math>5</math> white squares for each case, and the cases do not have any overlap.
 +
 +
So, the requested probability is <cmath>\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.</cmath>
 +
~apex304, TaeKim, MRENTHUSIASM
  
 +
==Solution 2==
  
−
==Solution 1==
+
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 <math>\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\textbf{(C) } \frac{1}{64}}</math>.
−
Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer.
 
  
−
There are <math>4</math> ways to choose the big diamond location from our <math>9</math> square grid.  From our given problem there are <math>4</math> different arrangements of triangles for every square. This implies that from having <math>1</math> diamond we are going to have <math>4^5</math> distinct patterns outside of the diamond. This gives a total of <math>4\cdot 4^5 = 4^6</math> favorable cases.
+
~aayr
  
 +
==Solution 3==
 +
Note that each tile must be in its precise place. Because of that, each diamond has a <math>\left(\frac14\right)^4</math> chance of appearing. And since there are 4 placements, our solution is<math>\frac{1}{4^4} \cdot 4 = \boxed{\textbf{(C) } \frac{1}{64}}</math>.
  
−
There are 9 squares and 4 possible designs for each square, giving <math>4^9</math> total outcomes. Thus, our desired probability is <math>\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \boxed{\text{(C)} \hspace{0.1 in} \dfrac{1}{64}}</math> .
+
~ligonmathkid2
−
-apex304 and TaeKim
 
  
−
==Solution 2 (Linearity of Expectation)==
+
==Solution 4 (Linearity of Expectation)==
−
Let <math>S_1, S_2, S_3</math>, and <math>S_4</math> denote the <math>4</math> smaller <math>2 \times 2</math> squares within the <math>3 \times 3</math> square in some order. For each <math>S_i</math>, let <math>X_i = 1</math> if it contains a large gray diamond tiling and <math>X_i = 0</math> otherwise. This means that <math>\mathbb{E}[X_i]</math> is the probability that square <math>S_i</math> has a large gray diamond, so <math>\mathbb{E}[X_1 + X_2 + X_3 + X_4]</math> is our desired probability. However, since there is only one possible way to arrange the squares within every <math>2 \times 2</math> square to form such a tiling, we have <math>\mathbb{E}[X_i] = (\tfrac{1}{4})^2 = \tfrac{1}{256}</math> for all <math>i</math> (as each of the smallest tiles has <math>4</math> possible arrangements), and from the linearity of expectation we get  
+
Let <math>S_1, S_2, S_3</math>, and <math>S_4</math> denote the <math>4</math> smaller <math>2 \times 2</math> squares within the <math>3 \times 3</math> square in some order. For each <math>S_i</math>, let <math>X_i = 1</math> if it contains a large gray diamond tiling and <math>X_i = 0</math> otherwise. This means that <math>\mathbb{E}[X_i]</math> is the probability that square <math>S_i</math> has a large gray diamond, so <math>\mathbb{E}[X_1 + X_2 + X_3 + X_4]</math> is our desired probability. However, since there is only one possible way to arrange the squares within every <math>2 \times 2</math> square to form such a tiling, we have <math>\mathbb{E}[X_i] = (\tfrac{1}{4})^4 = \tfrac{1}{256}</math> for all <math>i</math> (as each of the smallest tiles has <math>4</math> possible arrangements), and from the linearity of expectation we get  
−
<cmath>\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}}</cmath>
+
<cmath>\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}}.</cmath>
 
~eibc
 
~eibc
  
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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. <math>X_1 + X_2 + X_3 + X_4 \le 1</math>.
 
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. <math>X_1 + X_2 + X_3 + X_4 \le 1</math>.
 +
 
~numerophile
 
~numerophile
  
−
==Solution 3==
+
==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=PZJnyfpeDaEY_K2i&t=1070
 +
 
 +
~hsnacademy
 +
 
 +
==Video Solution by Math-X (Smart and Simple)==
 +
https://youtu.be/Ku_c1YHnLt0?si=wz7F8E_c7Hxv9GQ5&t=5137 ~Math-X
 +
 
 +
==Video Solution (THINKING CREATIVELY!!!)==
  
−
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 <math>\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\textbf{(C)}\ \frac{1}{64}}</math>
+
https://youtu.be/B5qcEnobmEU
  
−
~aayr
+
~Education, the Study of Everything
  
−
==Video Solution 1 by OmegaLearn (Using Cool Probability Technique)==
+
==Video Solution==
 
https://youtu.be/2t_Za0Y2IqY
 
https://youtu.be/2t_Za0Y2IqY
  
 
==Animated Video Solution==
 
==Animated Video Solution==
−
https://youtu.be/f4ffQEG0yUw
+
https://www.youtube.com/watch?v=wq5Ie6mUxvY&ab_channel=StarLeague
  
 
~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
Line 70: Line 109:
 
==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=2405
 
https://youtu.be/-N46BeEKaCQ?t=2405
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=3115
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/uJklHght9ds
 +
 +
~savannahsolver
 +
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s
 +
 +
~harungurcan
 +
 +
== Video Solution==
 +
 +
https://youtu.be/YFLSh4WoZrg
 +
~please like and subscribe
 +
 +
== Video Solution by Dr. David==
 +
 +
https://youtu.be/-KGQunxiVL4
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=22|num-a=24}}
 
{{AMC8 box|year=2023|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:36, 10 November 2024

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

Note that each tile must be in its precise place. Because of that, each diamond has a $\left(\frac14\right)^4$ chance of appearing. And since there are 4 placements, our solution is$\frac{1}{4^4} \cdot 4 = \boxed{\textbf{(C) } \frac{1}{64}}$.

~ligonmathkid2

Solution 4 (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 $\mathbb{E}[X_i] = (\tfrac{1}{4})^4 = \tfrac{1}{256}$ 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

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=PZJnyfpeDaEY_K2i&t=1070

~hsnacademy

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=wz7F8E_c7Hxv9GQ5&t=5137 ~Math-X

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/B5qcEnobmEU

~Education, the Study of Everything

Video Solution

https://youtu.be/2t_Za0Y2IqY

Animated Video Solution

https://www.youtube.com/watch?v=wq5Ie6mUxvY&ab_channel=StarLeague

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

Video Solution by Magic Square

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

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=3115

Video Solution by WhyMath

https://youtu.be/uJklHght9ds

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s

~harungurcan

Video Solution

https://youtu.be/YFLSh4WoZrg ~please like and subscribe

Video Solution by Dr. David

https://youtu.be/-KGQunxiVL4

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

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

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