|
|
| (One intermediate revision by one other user not shown) |
| Line 1: |
Line 1: |
| − | | + | #redirect[[2023 AMC 12B Problems/Problem 5]] |
| − | == Solution 1 ==
| |
| − | | |
| − | First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with <math>5</math> guesses, and one with <math>4</math>. Since the problem is asking for the minimum number, the answer is <math>\boxed{\text{(C) } 4}</math>.
| |
| − | | |
| − | ~yourmomisalosinggame (a.k.a. Aaron)
| |
| − | | |
| − | == Solution 2 ==
| |
| − | | |
| − | Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't used the adjacent condition, and then it clicks. If we eliminate the four squares with only one edge on the boundary of the 9x9 square. We are left with 5 diagonal squares, since our rectangle cant be diagonal, we can ensure that we find it in 4 moves. So our answer is :
| |
| − | <math>\boxed{\text{(C) } 4}</math>
| |
| − | | |
| − | ~arrowskyknight22
| |
