Difference between revisions of "2024 AMC 8 Problems/Problem 17"

Revision as of 16:48, 8 May 2024

1 Problem
2 Solution 1
3 Solution 2
4 Video Solution 1 (super clear!) by Power Solve
5 Video Solution 2 by Math-X (First understand the problem!!!)
6 Video Solution 3 by OmegaLearn.org
7 Video Solution 4 by SpreadTheMathLove
8 Video Solution by NiuniuMaths (Easy to understand!)
9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
10 Video Solution by Interstigation
11 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 $3$ x $3$ grid attacks all $8$ other squares, as shown below. Suppose a white king and a black king are placed on different squares of a $3$ x $3$ 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$", (1.5,1.5)); [/asy]$

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 32$

~Diagram by Andrei.martynau

Solution 1

If you place a king in any of the $4$ corners, they will have $5$ spots to go and there are $4$ corners, so $5 \times 4=20$ . If you place a king in any of the $3$ corners, they will have $3$ spots to go and there are $4$ sides so, $3 \times 4=12$ . That gives us $20+12=32$ spots to go into totally. So $\boxed{\textbf{(E)} 32}$ is the answer. ~andliu766

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 $\binom{4}{2}=6$

Corner-edge: For each corner, there are two edges that don't border it, $4\cdot2=8$

Edge-edge: The only possible combinations of this that work are top-bottom and left-right edges, so $2$ for this type

$6+8+2=16$

Multiply by two to account for arrangements of colors to get $\fbox{E) 32}$ ~ c_double_sharp

Video Solution 1 (super clear!) by Power Solve

https://youtu.be/SG4PRARL0TY

Video Solution 2 by Math-X (First understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=Q2e8OfkuzKZXmoau&t=4624

~Math-X

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

@@ Line 25: / Line 25: @@
 ==Solution 1==
-Corners 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>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>.
-Edges have <math>3</math> spots to go and there are <math>4</math> sides so, <math>3 \times 4=12</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>.
 That gives us <math>20+12=32</math> spots to go into totally.
 So <math>\boxed{\textbf{(E)} 32}</math> is the answer.

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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