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 | + | //diagram |
− | size( | + | 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 25: | Line 25: | ||
==Solution 1== | ==Solution 1== | ||
+ | |||
+ | We are given a 3x3 grid in the problem. Therefore, there are 9 possible spots for a triangle or circle to be placed, and we are then asked to choose any 3 possible positions in a row for either a triangle or circle. Thus we have <math>\binom{9}{3}</math> which is equivalent to <math>\boxed{\textbf{(D) }84}</math> | ||
+ | |||
+ | ~Clew28 | ||
+ | |||
+ | ==Solution 2 (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. | ||
Line 33: | Line 39: | ||
'''Case 2: 2 lines''': | '''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 39: | Line 45: | ||
~wamofan | ~wamofan | ||
− | ==Solution | + | ==Solution 3== |
− | We will only consider | + | 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== | ==Video Solution== | ||
Line 48: | Line 71: | ||
~Mathematical Dexterity | ~Mathematical Dexterity | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/Ij9pAy6tQSg?t=2250 | https://youtu.be/Ij9pAy6tQSg?t=2250 | ||
~Interstigation | ~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 10:02, 7 September 2024
Contents
Problem
A or is placed in each of the nine squares in a -by- grid. Shown below is a sample configuration with three s in a line. How many configurations will have three s in a line and three s in a line?
Solution 1
We are given a 3x3 grid in the problem. Therefore, there are 9 possible spots for a triangle or circle to be placed, and we are then asked to choose any 3 possible positions in a row for either a triangle or circle. Thus we have which is equivalent to
~Clew28
Solution 2 (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 of one shape and of another, so there are ways to arrange the lines and ways to pick which shape has only one line. In total, this is
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 ways to arrange the lines and ways to choose the last line. (We subtract 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
Finally, we add and multiply: .
~wamofan
Solution 3
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 ways to choose a column with all 's and ways to choose a column with all 's. The third column can be filled in ways. Therefore, we have a total of cases. However, we overcounted the cases with complete columns of with one symbol and complete column with another symbol. This happens in cases. . However, we have to remember to double our answer, giving us 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
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
~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.