Difference between revisions of "2024 AMC 8 Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a <math>3</math> x <math>3</math> grid attacks all <math>8</math> other squares, as shown below. Suppose a white king and a black king are placed on different squares of a <math>3</math> x <math>3</math> grid so that they do not attack each other. In how many ways can this be done? | + | A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a <math>3</math> x <math>3</math> grid attacks all <math>8</math> other squares, as shown below. Suppose a white king and a black king are placed on different squares of a <math>3</math> x <math>3</math> grid so that they do not attack each other (in other words, not right next to each other). In how many ways can this be done? |
− | ( | + | <asy> |
+ | /* AMC8 P17 2024, revised by Teacher David */ | ||
+ | unitsize(29pt); | ||
+ | import math; | ||
+ | add(grid(3,3)); | ||
+ | |||
+ | pair [] a = {(0.5,0.5), (0.5, 1.5), (0.5, 2.5), (1.5, 2.5), (2.5,2.5), (2.5,1.5), (2.5,0.5), (1.5,0.5)}; | ||
+ | |||
+ | for (int i=0; i<a.length; ++i) { | ||
+ | pair x = (1.5,1.5) + 0.4*dir(225-45*i); | ||
+ | draw(x -- a[i], arrow=EndArrow()); | ||
+ | } | ||
+ | |||
+ | label("$K$", (1.5,1.5)); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 32</math> | ||
+ | |||
+ | ~Diagram by Andrei.martynau | ||
==Solution 1== | ==Solution 1== | ||
− | + | If you place a king in any of the <math>4</math> corners, they will have <math>5</math> spots to go and there are <math>4</math> corners, so <math>5 \times 4=20</math>. | |
− | + | If you place a king in any of the <math>3</math> corners, they will have <math>3</math> spots to go and there are <math>4</math> sides so, <math>3 \times 4=12</math>. | |
− | 20+12=32 | + | That gives us <math>20+12=32</math> spots to go into totally. |
− | E ( | + | So <math>\boxed{\textbf{(E)} 32}</math> is the answer. |
+ | ~andliu766, captaindiamond868 (minor edits) | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We see that the center is not a viable spot for either of the kings to be in, as it would attack all nearby squares. | ||
+ | |||
+ | This gives three combinations: | ||
+ | |||
+ | Corner-corner: There are 4 corners, and none of them are touching orthogonally or diagonally, so it's <math>\binom{4}{2}=6</math> | ||
+ | |||
+ | Corner-edge: For each corner, there are two edges that don't border it, <math>4\cdot2=8</math> | ||
+ | |||
+ | Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so <math>2</math> for this type | ||
+ | |||
+ | |||
+ | <math>6+8+2=16</math> | ||
+ | |||
+ | Multiply by two to account for arrangements of colors to get <math>\fbox{E) 32}</math> ~ c_double_sharp | ||
+ | |||
+ | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
+ | https://youtu.be/5ZIFnqymdDQ?si=_SHs3ZXS1ZKF4-v2&t=2381 | ||
+ | |||
+ | ~hsnacademy | ||
==Video Solution 1 by Math-X (First understand the problem!!!)== | ==Video Solution 1 by Math-X (First understand the problem!!!)== | ||
− | https://youtu.be/ | + | https://youtu.be/BaE00H2SHQM?si=Q2e8OfkuzKZXmoau&t=4624 |
~Math-X | ~Math-X | ||
+ | |||
==Video Solution 2 (super clear!) by Power Solve== | ==Video Solution 2 (super clear!) by Power Solve== | ||
https://youtu.be/SG4PRARL0TY | https://youtu.be/SG4PRARL0TY | ||
+ | |||
+ | ==Video Solution 3 by OmegaLearn.org== | ||
+ | https://youtu.be/UJ3esPnlI5M | ||
+ | |||
+ | ==Video Solution 4 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=Svibu3nKB7E | ||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
+ | |||
+ | https://www.youtube.com/watch?v=quWFZIahQCg | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=1922 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2024|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:01, 28 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution (A Clever Explanation You’ll Get Instantly)
- 5 Video Solution 1 by Math-X (First understand the problem!!!)
- 6 Video Solution 2 (super clear!) by Power Solve
- 7 Video Solution 3 by OmegaLearn.org
- 8 Video Solution 4 by SpreadTheMathLove
- 9 Video Solution by NiuniuMaths (Easy to understand!)
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 See Also
Problem
A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a x
grid attacks all
other squares, as shown below. Suppose a white king and a black king are placed on different squares of a
x
grid so that they do not attack each other (in other words, not right next to each other). In how many ways can this be done?
~Diagram by Andrei.martynau
Solution 1
If you place a king in any of the corners, they will have
spots to go and there are
corners, so
.
If you place a king in any of the
corners, they will have
spots to go and there are
sides so,
.
That gives us
spots to go into totally.
So
is the answer.
~andliu766, captaindiamond868 (minor edits)
Solution 2
We see that the center is not a viable spot for either of the kings to be in, as it would attack all nearby squares.
This gives three combinations:
Corner-corner: There are 4 corners, and none of them are touching orthogonally or diagonally, so it's
Corner-edge: For each corner, there are two edges that don't border it,
Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so for this type
Multiply by two to account for arrangements of colors to get ~ c_double_sharp
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=_SHs3ZXS1ZKF4-v2&t=2381
~hsnacademy
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=Q2e8OfkuzKZXmoau&t=4624
~Math-X
Video Solution 2 (super clear!) by Power Solve
Video Solution 3 by OmegaLearn.org
Video Solution 4 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=quWFZIahQCg
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1922
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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.