AMC12 2023 B

AMC12 2023 B · Q5

AMC12 2023 B · Q5. It mainly tests Pigeonhole principle, Casework.

You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

你正在玩一个游戏。一个 $2 \times 1$ 矩形覆盖 $3 \times 3$ 方格网格中两个相邻的方格（可以水平或垂直放置），但你不知道覆盖了哪两个方格。你的目标是找到至少一个被矩形覆盖的方格。一“回合”是你猜测一个方格，然后被告知该方格是否被隐藏的矩形覆盖。为了确保至少有一个猜测的方格被矩形覆盖，你需要的最少回合数是多少？

(A) 3 3

(B) 5 5

(D) 8 8

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles. Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$. Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$

注意 $3\times3$ 方格网格总共有 $12$ 个可能的 $2\times1$ 矩形。假设你第一回合选择中间方格。中间方格被 $4$ 个矩形覆盖，其余 $8$ 个方格每个最多被 $2$ 个未计的矩形覆盖。这意味着回合数至少是 $1+\frac{12-4}{2}=1+4=5$ 。现在假设你不选择中间方格。边中方格最多被 $3$ 个未计矩形覆盖，角方格最多被 $2$ 个。在这种情况下，覆盖所有 $12$ 个矩形所需的最少回合数是 $12\div 3=4$ 。为了证明只选择边方格确实覆盖所有 $12$ 个矩形，我们需要显示每个边方格覆盖的 $3$ 个矩形不重叠。画出覆盖一个方格的矩形，看到它们形成 T 形，且不覆盖其他边方格。因此，答案是 $4$ 。

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