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Difference between revisions of "2022 AMC 8 Problems/Problem 23"

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A <math>\triangle</math> or <math>\bigcirc</math> is placed in each of the nine squares in a <math>3</math>-by-<math>3</math> grid. Shown below is a sample configuration with three <math>\triangle</math>s in a line.
 
A <math>\triangle</math> or <math>\bigcirc</math> is placed in each of the nine squares in a <math>3</math>-by-<math>3</math> grid. Shown below is a sample configuration with three <math>\triangle</math>s in a line.
 
<asy>
 
<asy>
−
//diagram by kante314
+
//diagram
−
size(3.3cm);
+
size(5cm);
−
defaultpen(linewidth(1));
+
defaultpen(linewidth(1.5));
 
real r = 0.37;
 
real r = 0.37;
 
path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle;
 
path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle;
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<math>\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96</math>
 
<math>\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96</math>
  
−
==Solution 1==
+
==Solution 1 (Casework)==
 
Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.
 
Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.
  
 
We take casework:
 
We take casework:
  
−
<i>Case 1: 3 lines</i>
+
'''Case 1: 3 lines''':
 
In this case, the lines would need to be <math>2</math> of one shape and <math>1</math> of another, so there are <math>\frac{3!}{2} = 3</math> ways to arrange the lines and <math>2</math> ways to pick which shape has only one line. In total, this is <math>3\cdot 2 = 6.</math>
 
In this case, the lines would need to be <math>2</math> of one shape and <math>1</math> of another, so there are <math>\frac{3!}{2} = 3</math> ways to arrange the lines and <math>2</math> ways to pick which shape has only one line. In total, this is <math>3\cdot 2 = 6.</math>
  
−
<i>Case 2: 2 lines</i>
+
'''Case 2: 2 lines''':
−
In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are <math>3! = 6</math> ways to arrange the lines and <math>2^3-2 = 6</math> ways to choose the last line. In total, this is <math>6\cdot 6 = 36.</math>
+
In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are <math>3! = 6</math> ways to arrange the lines and <math>2^3-2 = 6</math> ways to choose the last line. (We subtract <math>2</math> from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is <math>6\cdot 6 = 36.</math>
  
 
Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math>.
 
Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math>.
Line 40: Line 40:
  
 
==Solution 2==
 
==Solution 2==
−
We will only consider columns, but at the end our answer should be multiplied by <math>2</math>. There are <math>3</math> ways to choose a column for <math>\bigcirc</math> and <math>2</math> ways to choose a column for <math>\triangle</math>. The third column can be filled in <math>2^3=8</math> ways. Therefore, we have <math>3\cdot2\cdot8=48</math> ways. However, we overcounted the cases with <math>2</math> complete columns of with one symbol and <math>1</math> complete columns with another symbol. This happens in <math>2\cdot3=6</math> ways. <math>48-6=42</math>. However, we have to remember to double our answer giving us <math>\boxed{\textbf{(D) }84}</math>.
+
We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are <math>3</math> ways to choose a column with all <math>\bigcirc</math>'s and <math>2</math> ways to choose a column with all <math>\triangle</math>'s. The third column can be filled in <math>2^3=8</math> ways. Therefore, we have a total of <math>3\cdot2\cdot8=48</math> cases. However, we overcounted the cases with <math>2</math> complete columns of with one symbol and <math>1</math> complete column with another symbol. This happens in <math>2\cdot3=6</math> cases. <math>48-6=42</math>. However, we have to remember to double our answer, giving us <math>\boxed{\textbf{(D) }84}</math> ways to complete the grid.
  
 
~MathFun1000
 
~MathFun1000
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
 +
I made A SECOND VERSION ( very easy to understand)
 +
https://youtu.be/ukCWuMbxxLU
 +
 +
~Math-X
 +
 +
==Video Solution==
 +
https://youtu.be/p7UHadjWqLg
 +
 +
Please like and subscribe!
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/fL7DKXZjmAo?t=239
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=or4pKVzQ3gI
 +
 +
~Mathematical Dexterity
 +
 +
==Video Solution==
 +
https://youtu.be/Ij9pAy6tQSg?t=2250
 +
 +
~Interstigation
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=KYglbGTvfsY 
 +
 +
~David
 +
 +
==Video Solution==
 +
https://youtu.be/0orAAUaLIO0?t=257
 +
 +
~STEMbreezy
 +
 +
==Video Solution==
 +
https://youtu.be/YYvbTopjB1E
 +
 +
~savannahsolver
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=22|num-a=24}}
 
{{AMC8 box|year=2022|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:14, 19 November 2024

Problem

A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$-by-$3$ grid. Shown below is a sample configuration with three $\triangle$s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

Solution 1 (Casework)

Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.

We take casework:

Case 1: 3 lines: In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$

Case 2: 2 lines: In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$

Finally, we add and multiply: $2(36+6)=2(42)=\boxed{\textbf{(D) }84}$.

~wamofan

Solution 2

We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$'s and $2$ ways to choose a column with all $\triangle$'s. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$. However, we have to remember to double our answer, giving us $\boxed{\textbf{(D) }84}$ ways to complete the grid.

~MathFun1000

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

I made A SECOND VERSION ( very easy to understand) https://youtu.be/ukCWuMbxxLU

~Math-X

Video Solution

https://youtu.be/p7UHadjWqLg

Please like and subscribe!

Video Solution by OmegaLearn

https://youtu.be/fL7DKXZjmAo?t=239

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=or4pKVzQ3gI

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=2250

~Interstigation

Video Solution

https://www.youtube.com/watch?v=KYglbGTvfsY

~David

Video Solution

https://youtu.be/0orAAUaLIO0?t=257

~STEMbreezy

Video Solution

https://youtu.be/YYvbTopjB1E

~savannahsolver

See Also

2022 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

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