AMC12 2023 A

AMC12 2023 A · Q21

AMC12 2023 A · Q21. It mainly tests Probability (basic), Geometry misc.

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$?

如果 $A$ 和 $B$ 是多面体的顶点，定义距离 $d(A,B)$ 为连接 $A$ 和 $B$ 所需穿越的多面体边的最小数量。例如，如果 $\overline{AB}$ 是多面体的边，则 $d(A, B) = 1$，但如果 $\overline{AC}$ 和 $\overline{CB}$ 是边而 $\overline{AB}$ 不是边，则 $d(A, B) = 2$。让 $Q$、$R$ 和 $S$ 是从正二十面体（由 20 个等边三角形组成的正多面体）中随机选择的不同的顶点。$d(Q, R) > d(R, S)$ 的概率是多少？

(A) \frac{7}{22} \frac{7}{22}

(B) \frac{1}{3} \frac{1}{3}

(D) \frac{5}{12} \frac{5}{12}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (A)

正确答案：（A）

Solution

To find the total amount of vertices we first find the amount of edges, and that is $\frac{20 \times 3}{2}$. Next, to find the amount of vertices we can use Euler's characteristic, $V - E + F = 2$, and therefore the amount of vertices is $12$ So there are $\binom{12}{3} = 220$ ways to choose 3 distinct points. Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of $1 \leq d(Q, R), d(R, S) \leq 3$ With some case work, we get two cases: Case 1: $d(Q, R) = 3; d(R, S) = 1, 2$ Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S. Then, we get $12 \times 1 \times 10 = 120$ (ways to choose R × ways to choose Q × ways to choose S) Case 2: $d(Q, R) = 2; d(R, S) = 1$ We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row. Therefore, we have $12 \times 5 \times 5 = 300$ (ways to choose R × ways to choose Q × ways to choose S) In total, we have $420$ ways, but we must divide by $3!$ to account for permutations, giving us $70/220 = \boxed{\textbf{(A) } 7/22}.$

首先，计算顶点总数。我们先求边的数量，为 $\frac{20 \times 3}{2}$。然后，使用欧拉特征式 $V - E + F = 2$，因此顶点数为 $12$。因此，选择 3 个不同点的方案数为 $\binom{12}{3} = 220$。在二十面体中，一点到另一点的最大距离为 3，因此 $1 \leq d(Q, R), d(R, S) \leq 3$。通过分类讨论，有两种情况：情况 1: $d(Q, R) = 3$；$d(R, S) = 1, 2$ 对于 $Q$，只有与 $R$ 对顶的一个点，因此 $Q$ 只有 1 种选择，$S$ 可以是其他任意点。因此，有 $12 \times 1 \times 10 = 120$ 种方式（选择 $R$ 的方式 × 选择 $Q$ 的方式 × 选择 $S$ 的方式）。情况 2: $d(Q, R) = 2$；$d(R, S) = 1$ 可以将二十面体可视化为 4 行：第一行 1 个顶点，第二行 5 个，第三行 5 个，第四行 1 个。将 $R$ 设为第一行的顶点，有 12 种选择。然后，$Q$ 可以是第三行的 5 个点之一，$S$ 可以是第二行的 5 个点之一。因此，有 $12 \times 5 \times 5 = 300$ 种方式。总计 $420$ 种方式，但需除以 $3!$ 以考虑排列，得到 $70/220 = \boxed{\textbf{(A) } 7/22}$。

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