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 "2022 AMC 8 Problems/Problem 23"

(added solution 2)
(→‎Video Solution)
 
(52 intermediate revisions by 27 users not shown)
Line 2: Line 2:
 
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;
Line 24: Line 24:
 
<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==
+
==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 2 of one shape and 1 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>.
  
 
~wamofan
 
~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 <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
 +
 +
==Video Solution (An excellent video solution that you will pick up)==
 +
https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=2978
 +
~hsnacademy
 +
 +
==Video Solution by Math-X==
 +
 +
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
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/CRHFEwZeqEM
  
 
==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 21:14, 18 January 2025

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 (An excellent video solution that you will pick up)

https://youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=2978 ~hsnacademy

Video Solution by Math-X

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

Video Solution by Dr. David

https://youtu.be/CRHFEwZeqEM

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_23&oldid=240540"