Difference between revisions of "2022 AMC 8 Problems/Problem 23"
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==Solution== | ==Solution== | ||
| + | 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: | ||
| + | |||
| + | <i>Case 1: 3 lines</i> | ||
| + | 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> | ||
| + | |||
| + | <i>Case 2: 2 lines</i> | ||
| + | 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> | ||
| + | |||
| + | Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math> | ||
| + | |||
| + | ~wamofan | ||
==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}} | ||
Revision as of 20:15, 28 January 2022
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.
![[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]](http://latex.artofproblemsolving.com/f/7/9/f79f74a5bed66611fe027a9dd5ab0948b3a033ac.png)
How many configurations will have three s in a line and three s in a line?
Solution
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 
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. In total, this is 
Finally, we add and multiply: 
~wamofan
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. 
