2023 AMC 12A Problems/Problem 21

Problem

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 $d(Q, R) > d(R, S)$ ?

Solution 1

First, note that a regular icosahedron has 12 vertices. So there are ${}_{12}P_{3} = 1320$ ways to choose 3 distinct points. Now, the furthest distance we can get from one point to another point in a icosahedron is 3. Which gives us a range of $1 ≤ d(Q, R), d(R, S) ≤ 3$ (Error compiling LaTeX. Unknown error_msg) With some case work, we get: Case 1: $d(Q, R)=3; d(R, S)=1,2$ $12×1×10=120$ Case 2: $d(Q, R)=2; d(R, S)=1$ $12×5×5=300$ Hence, $P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \frac{7}{22}$

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2023 AMC 12A Problems/Problem 21

Problem

Solution 1

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