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; | ||
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~wamofan | ~wamofan | ||
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==Solution 2== | ==Solution 2== | ||
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==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== | ||
− | https://youtu.be/ | + | |
+ | I made A SECOND VERSION ( very easy to understand) | ||
+ | https://youtu.be/ukCWuMbxxLU | ||
~Math-X | ~Math-X | ||
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~savannahsolver | ~savannahsolver | ||
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==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
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 (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 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 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.