2023 AIME II 第 7 题

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7.

1212 边形的每个顶点都要涂成红色或蓝色,因此共有 2122^{12} 种涂色。求其中满足如下性质的涂色数:不存在四个同色顶点恰好是一个矩形的四个顶点。

Each vertex of a regular dodecagon (1212-gon) is to be colored either red or blue, and thus there are 2122^{12} possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.

答案:928
知识点:有限制的排列乘法原理分类讨论
难度评级:2600
解答:

十二个顶点在同一个圆上,而圆内接矩形的对角线必须经过圆心。因此这些顶点构成的矩形恰好对应于两条不同的直径, 即从 66 对对顶顶点中选两对。出现同色矩形,当且仅当有两对对顶顶点都被同一种颜色涂满。

每对对顶顶点独立地可以是全红(11 种)、全蓝(11 种)或混合(22 种)。一种涂色有效,当且仅当全红的对数至多为一,全蓝的对数也至多为一。按全红对数和全蓝对数计数: 26+625+625+6524=64+192+192+480=928. \begin{gathered} 2^6 + 6 \cdot 2^5 \\ {}+ 6 \cdot 2^5 + 6 \cdot 5 \cdot 2^4 \\ = 64 + 192 + 192 + 480 \\ = 928. \end{gathered}

The twelve vertices lie on a circle, and a rectangle inscribed in a circle must have its diagonals pass through the center. So the rectangles with vertices among the twelve are exactly the pairs of distinct diameters, where the diameters join the 66 antipodal pairs of vertices. A monochromatic rectangle appears exactly when two antipodal pairs are each colored solidly in the same color.

Each antipodal pair is independently both red (11 way), both blue (11 way), or mixed (22 ways). A coloring is valid exactly when at most one pair is both red and at most one pair is both blue. Counting by the numbers of solid red and solid blue pairs: 26+625+625+6524=64+192+192+480=928. \begin{gathered} 2^6 + 6 \cdot 2^5 \\ {}+ 6 \cdot 2^5 + 6 \cdot 5 \cdot 2^4 \\ = 64 + 192 + 192 + 480 \\ = 928. \end{gathered}

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