2013 AMC 10A Problem 25

Number the corners $1$ to $8$ around the octagon.

1. When do two diagonals cross inside? Pick any $4$ corners, say $1, 3, 4, 7$. Of the diagonals using these corners, exactly one pair crosses inside: the one joining opposite corners of the four-sided shape, here $1\text{-}4$ and $3\text{-}7$. And every crossing pair comes from its $4$ endpoints this way.
2. Count pairs of crossing diagonals. The number of ways to choose $4$ of the $8$ corners is
$$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70.$$
So $70$ pairs of diagonals cross. If every crossing point had only $2$ diagonals through it, the answer would be $70$. But some points have more, and those were counted more than once.
3. The center. The four longest diagonals $1\text{-}5$, $2\text{-}6$, $3\text{-}7$, $4\text{-}8$ all go through the center. Four diagonals make $6$ pairs, so the center was counted $6$ times instead of once. Subtract $5$.
4. Other points where three meet. Fold the octagon along the diagonal $1\text{-}5$. Corner $2$ lands on $8$, $3$ on $7$, $4$ on $6$. So diagonal $2\text{-}7$ folds onto $8\text{-}3$. These two mirror images cross each other, and a crossing of two mirror images must lie on the fold line. So $2\text{-}7$, $3\text{-}8$ and $1\text{-}5$ all pass through one point.
The same happens with $4\text{-}7$ and $3\text{-}6$ on the other side of the center. So each long diagonal has $2$ such points, and there are $4$ long diagonals: $8$ points.
5. Fix those 8 points. Three diagonals make $3$ pairs, so each point was counted $3$ times instead of once. Subtract $2$ for each: $8 \times 2 = 16$.
6. Answer. There are no other spots where three or more diagonals meet, so
$$70 - 5 - 16 = 49.$$

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

"Choose $4$ corners, get one crossing" is the standard way to count where diagonals meet, and it is exact when no three diagonals share a point. Regular polygons are very symmetric, and symmetry forces extra meeting points: two diagonals that are mirror images meet on the mirror line. For a point where $k$ diagonals meet, correct the count by subtracting (number of pairs among the $k$) minus $1$.

You can also narrow down the choices. The count $70$ assumes no three diagonals meet, but the four long diagonals share the center, which alone brings the count down to $70 - 5 = 65$. So (C), (D) and (E) are too big. Choice (B) $65$ is what you get if you find the center but miss the points of Step 4. Finding even one of those points pushes the count below $65$, which leaves only (A).