Question

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 $AB$ is an edge of the polyhedron, then $d(A, B) = 1,$ but if $AC$ and $CB$ are edges and $AB$ is not an edge, then $d(A, B) = 2.$ Let $Q,$ $R,$ and $S$ be randomly chosen distinct vertices of a regular icosahedron (a regular polyhedron made up of $20$ equilateral triangles). What is the probability that $d(Q, R) \gt d(R, S)?$

$\dfrac{7}{22}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

Solution:

Fix $R.$ Of the other $11$ vertices, $5$ sit at distance $1,$ $5$ at distance $2,$ and $1$ (the opposite vertex) at distance $3.$ Pick ordered distinct $Q, S$ from these $11:$ that's $11 \cdot 10 = 110$ pairs. The ones with $d(R,Q) = d(R,S)$ number $5\cdot4 + 5\cdot4 + 1\cdot0 = 40,$ so $P(\text{equal}) = \frac{40}{110} = \frac{4}{11}.$ By the symmetry between $Q$ and $S,$ the $\gt$ and $\lt$ cases split the rest evenly, so $P(d(Q,R) \gt d(R,S)) = \frac{1 - \frac4{11}}{2} = \frac{7}{22}.$ Thus, A is the correct answer.

2023 AMC 10A Problem 25

Solution:

Problem 25 in Other Years