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

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~MathFun1000
 
~MathFun1000
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== Video Solution by OmegaLearn ==
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https://youtu.be/fL7DKXZjmAo?t=239
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~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
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~Mathematical Dexterity
 
~Mathematical Dexterity
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/Ij9pAy6tQSg?t=2250
 
https://youtu.be/Ij9pAy6tQSg?t=2250

Revision as of 02:52, 16 January 2023

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 by kante314 size(3.3cm); defaultpen(linewidth(1)); 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. 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 columns, but at the end our answer should be multiplied by $2$. There are $3$ ways to choose a column for $\bigcirc$ and $2$ ways to choose a column for $\triangle$. The third column can be filled in $2^3=8$ ways. Therefore, we have $3\cdot2\cdot8=48$ ways. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ ways. $48-6=42$. However, we have to remember to double our answer giving us $\boxed{\textbf{(D) }84}$.

~MathFun1000

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

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

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

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