AMC10 2023 A

AMC10 2023 A · Q18

AMC10 2023 A · Q18. It mainly tests Basic counting (rules of product/sum), Geometry misc.

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 meet?

菱形十二面体是一个有 $12$ 个全等菱形面的立体图形。在每个顶点，有 $3$ 或 $4$ 条边相交，取决于顶点。有多少个顶点恰好有 $3$ 条边相交？

(A) 5 5

(B) 6 6

(D) 8 8

(E) 9 9

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$. There are $12$ faces. There are $24$ edges, because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $2-12+24=14$ vertices. Now note that the sum of the degrees of all the points is $48$(the number of edges times 2). Let $x=$ the number of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$. Solving this equation gives $x = \boxed{\textbf{(D) }8}$. ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)

注意欧拉公式 $\text{顶点数}+\text{面数}-\text{边数}=2$。有 $12$ 个面。有 $24$ 条边，因为 $12$ 个面各有四条边，每条边被两个面共享。现在知道有 $2-12+24=14$ 个顶点。现在注意所有点的度数和为 $48$（边数乘以2）。设 $x=$ 有 $3$ 条边的顶点数。现在有 $\frac{3x+4(14-x)}{2}=24$。解此方程得 $x = \boxed{\textbf{(D)}8}$。

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