AMC12 2025 B

AMC12 2025 B · Q13

AMC12 2025 B · Q13. It mainly tests Casework, Counting with symmetry / Burnside (rare).

A circle has been divided into 6 sectors of different sizes. Then 2 of the sectors are painted red, 2 painted green, and 2 painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below. How many different colorings are possible?

一个圆被分成6个不同大小的扇形。然后其中2个扇形涂成红色，2个涂成绿色，2个涂成蓝色，使得没有两个相邻扇形涂成相同颜色。下面展示了一种这样的着色。有多少种不同的着色方式？

(A) 12 12

(B) 16 16

(D) 24 24

(E) 28 28

Answer

Correct choice: (D)

正确答案：（D）

Solution

Create an arbitrary six slice circular diagram. The first slice has 3 options, and the second has 2, third has 2, fourth has 2, fifth has 2, and the sixth has only 1 color to be chosen. Understand that if X is the set containing all possible cases of the circle, then X has a correlation to another set, call it K, which contains all the colors. This can be quite easily sought as a rotation of one element in X that accurately maps to another element in X is considered the same. In simpler terms, to account for overlapping cases in X, one must divide by 2 (as two colorings that are rotated are considered different, and we don't want that as it defeats the purpose of "unique"). Then, multiplying out we have 3×24=48. Dividing by 2 to account for overlap in X, we get 48/2 or $\boxed{\textbf{(D) } 24}$

画一个任意的六扇形圆形图。第一扇形有3种选择，第二种有2种，第三种有2种，第四种有2种，第五种有2种，第六种只有1种颜色可选。理解如果$X$是包含所有可能圆形情况的集合，则$X$与另一个集合$K$相关，其中$K$包含所有颜色。这可以通过旋转$X$中的一个元素精确映射到$X$中的另一个元素来轻松求得，视为相同。简单来说，为了考虑$X$中的重叠情况，必须除以2（因为两个旋转的着色被视为不同，但我们不希望这样，因为这违背了“独特”的目的）。然后，乘积为$3\times24=48$。除以2以考虑$X$中的重叠，得$48/2$或$\boxed{\textbf{(D) }24}$

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