2013 AMC 10B Problem 12

1. Count the segments. A pentagon has $5$ sides. Each corner connects by a diagonal to the $2$ corners that are not its neighbors. That is $5 \times 2 = 10$ diagonal ends, and each diagonal has $2$ ends, so there are $5$ diagonals. In total there are $10$ segments.
2. Count the lengths. The pentagon is regular, so turning it around one corner's worth matches it with itself. That means all $5$ sides have one length and all $5$ diagonals have another (longer) length.
3. Count all pairs. Choosing $2$ of the $10$ segments: $\frac{10 \times 9}{2} = 45$ pairs.
4. Count matching pairs. Two sides: $\frac{5 \times 4}{2} = 10$ pairs. Two diagonals: also $10$ pairs. Total $20$.
5. The probability is $\frac{20}{45} = \frac{4}{9}$.

The answer is $\boxed{\textbf{(B)}\ \frac{4}{9}}$.

The symmetry of the regular pentagon turns ten segments into just two groups of five. "Same length" becomes "same group", which is simple counting. Look for symmetry to sort objects into groups before you count.

Pick the segments one at a time. Whatever the first segment is, it belongs to a group of $5$. Of the $9$ segments left, $4$ are in the same group. So the chance the second one matches is $\frac{4}{9}$.

A regular pentagon has $5$ sides, all of one length, and $\binom{5}{2} - 5 = 5$ diagonals, all of another length (by the pentagon's rotational symmetry). So $S$ has $10$ segments in two length classes of $5$ each.

The number of ways to pick two segments is $\binom{10}{2} = 45$. The pair has equal lengths exactly when both are sides or both are diagonals: $\binom{5}{2} + \binom{5}{2} = 10 + 10 = 20$ pairs.

The probability is $\frac{20}{45} = \frac{4}{9}$.

The answer is $\boxed{\textbf{(B)}\ \frac{4}{9}}$.