2023 AMC 10A Problem 25

1. Fix $R$ first. Every vertex of the icosahedron looks the same, so it does not matter which vertex $R$ is. What matters is how far $Q$ and $S$ are from $R$.
2. Sort the other vertices by distance from $R$. The icosahedron has $12$ vertices, and $5$ edges meet at each one.
- Distance $1$: the $5$ neighbors of $R$.
- Distance $3$: the $1$ vertex directly opposite $R$ (at the far end of the solid).
- Distance $2$: everything else, $12 - 1 - 5 - 1 = 5$ vertices. (Picture the icosahedron as a top vertex, a ring of $5$, a second ring of $5$, and a bottom vertex.)

3. Count all choices of $Q$ and $S$. They are two different vertices from the other $11$, in order: $11 \times 10 = 110$ ways.
4. Count the ways $Q$ is farther than $S$.
- $Q$ at distance $2$, $S$ at distance $1$: $5 \times 5 = 25$ ways.
- $Q$ at distance $3$, $S$ at distance $1$ or $2$: $1 \times 10 = 10$ ways.

Total: $25 + 10 = 35$ ways.
5. Probability:
$$\frac{35}{110} = \frac{7}{22}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{7}{22}}$.

A hard-looking 3D question turns into a table once you stand at one vertex and sort the rest by distance: $5, 5, 1$. After that it is just counting ordered pairs. It also helps to notice that $Q$ and $S$ play the same role, so "$Q$ farther" and "$S$ farther" are equally likely, and the only reason the answer is not $\frac12$ is the chance of a tie.

Use that symmetry directly. Count the ties: both $Q$ and $S$ at distance $1$ gives $5 \times 4 = 20$ ordered pairs, both at distance $2$ gives another $20$, and both at distance $3$ is impossible (only one vertex). So a tie has probability $\frac{40}{110} = \frac{4}{11}$. The remaining $\frac{7}{11}$ is split equally between "$Q$ farther" and "$S$ farther," so the answer is $\frac12 \times \frac{7}{11} = \frac{7}{22}$.