2013 AMC 10A Problem 25

Two diagonals cross inside the octagon exactly when their four endpoints alternate around the boundary, and any $4$ vertices determine exactly one such crossing pair. So there are $\binom{8}{4} = 70$ crossing pairs of diagonals. If no three diagonals met at a point, the answer would be $70$; in a regular octagon some do, and those points are over-counted.

Label the vertices $1$ through $8$ in order.

The center. The four long diagonals $1\text{-}5$, $2\text{-}6$, $3\text{-}7$, $4\text{-}8$ all pass through the center. Those account for $\binom{4}{2} = 6$ crossing pairs but only $1$ point: subtract $5$.

Triple points. Take a long diagonal, say $1\text{-}5$, and the two diagonals $2\text{-}7$ and $3\text{-}8$, which are mirror images of each other across the line $1\text{-}5$. Mirror-image segments cross on the mirror line, and since $2\text{-}7$ and $3\text{-}8$ do cross (their endpoints alternate: $2, 3, 7, 8$), all three diagonals pass through one point. Each long diagonal carries two such points (one on each side of the center, using $\{2\text{-}7, 3\text{-}8\}$ and $\{4\text{-}7, 3\text{-}6\}$ for $1\text{-}5$), giving $4 \cdot 2 = 8$ triple points. Each is counted $\binom{3}{2} = 3$ times but is $1$ point: subtract $2$ per point, $16$ in all.

No other concurrences occur. Therefore the number of distinct interior intersection points is
$$70 - 5 - 16 = 49.$$

The answer is $\boxed{\textbf{(A)}\ 49}$.

The bijection "$4$ vertices $\leftrightarrow$ one crossing pair" is the standard way to count diagonal intersections, and $\binom{n}{4}$ is exact for polygons in general position. Regular polygons break general position through symmetry: mirror-image diagonals must meet on the axis, which is where the extra concurrences come from. The correction for a point where $k$ diagonals meet is $\binom{k}{2} - 1$.

Count by direct drawing along one long diagonal: on $1\text{-}5$ the interior points, in order, are the two triple points and the center, plus ordinary crossings with the remaining diagonals. Combined with the octagon's $8$-fold symmetry this reproduces $49$, but it is easy to lose track; the $\binom{8}{4}$ correction method is more reliable. Choice (C) $70$ is the general-position count and can be ruled out immediately once you notice the four long diagonals share the center.

• Answering 70 by assuming no three diagonals are concurrent; the regular octagon has concurrency at the center and at eight other points.
• Fixing only the center ($70 - 5 = 65$, choice B) and missing the eight triple points.