Question

菱形十二面体是一个有 $12$ 个全等菱形面的立体。在每个顶点处，根据顶点不同，会有 $3$ 条或 $4$ 条棱相交。恰有 $3$ 条棱相交的顶点有多少个？

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meeting?

$5$

$6$

$7$

$8$

$9$

答案：D

知识点：欧拉多面体公式多面体图论双重计数

难度评级：1660

解答：

每个菱形有 $4$ 条边，每条棱由 $2$ 个面共享，所以 $E = \frac{12 \cdot 4}{2} = 24$ 。已知 $F = 12$ ，由欧拉公式得 $V = 2 - F + E = 14$ 。设 $x$ 个顶点有 $3$ 条棱相交，其余 $14 - x$ 个顶点有 $4$ 条棱相交。度数之和等于棱数的两倍： $3x + 4(14 - x) = 2E = 48$ ，所以 $x = 8$ 。因此，答案是 D。

Each rhombus has $4$ edges, and every edge is shared by $2$ faces, so $E = \frac{12 \cdot 4}{2} = 24.$ With $F = 12,$ Euler's formula gives $V = 2 - F + E = 14.$ Suppose $x$ vertices have $3$ edges and the other $14 - x$ have $4.$ The degrees sum to twice the edge count: $3x + 4(14 - x) = 2E = 48,$ so $x = 8.$ Therefore, the answer is D.

2023 AMC 10A 第 18 题

答案：D

难度评级：1660

解答：

其他年份的第 18 题