AMC10 2013 A

Answer (A): Label the octagon $ABCDEFGH$. There are 20 diagonals in all, 5 with endpoints at each vertex. The diagonals are of three types: • Diagonals that skip over only one vertex, such as $AC$ or $AG$. These diagonals intersect with each of the five diagonals with endpoints at the skipped vertex. • Diagonals that skip two vertices, such as $\overline{AD}$ or $\overline{AF}$. These diagonals intersect with four of the five diagonals that have endpoints at each of the two skipped vertices. • Diagonals that cross to the opposite vertex, such as $\overline{AE}$. These diagonals intersect with three of the five diagonals that have endpoints at each of the three skipped vertices. Therefore, from any given vertex, the diagonals will intersect other diagonals at $2\cdot 5+2\cdot 8+1\cdot 9=35$ points. Counting from all 8 vertices, the total is $8\cdot 35=280$ points. Observe that, by symmetry, all four diagonals that cross to the opposite vertex intersect in the center of the octagon. This single intersection point has been counted 24 times, 3 from each of the 8 vertices. Further observe that at each of the vertices of the smallest internal octagon created by the diagonals, 3 diagonals intersect. For example, $\overline{AD}$ intersects with $\overline{CH}$ on $\overline{BF}$. These 8 intersection points have each been counted 12 times, 2 from each of the 6 affected vertices. The remaining intersection points each involve only two diagonals and each has been counted 4 times, once from each endpoint. These number $\dfrac{280-24-8\cdot 12}{4}=40$. There are therefore $1+8+40=49$ distinct intersection points in the interior of the octagon.