A regular dodecagon has consecutive vertices $C_{1}, C_{2}, \ldots, C_{12}$. The even-indexed vertices are connected to every vertex. How many internal points do these diagonals intersect and into how many parts do they divide this dodecagon?
The figure shows a $60^{\circ}$ rotational symmetry. The $S_{12}$ dodecagon, when rotated by any multiple of $60^{\circ}$ around its center $O$, maps every even-indexed vertex to another even-indexed vertex and every odd-indexed vertex to another odd-indexed vertex. Every drawn diagonal maps to another drawn diagonal. Therefore, it is sufficient to count the sub-regions and diagonal intersection points within the kite $C_{12} C_{1} C_{2} O=N$, including the boundary segment $O C_{12}$ in $N$, but not including the boundary segment $O C_{2}$ and the point $O$ itself. It is evident that the number of sub-regions in $S_{12}$ is 6 times the number in $N$, and the number of intersection points in $S_{12}$ is 1 more than 6 times the number in $N$.
![](https://cdn.mathpix.com/cropped/2024_05_02_c9de918ad928e3fde173g-1.jpg?height=648&width=1114&top_left_y=431&top_left_x=510)

In the figure, two diagonals symmetric about the $C_{12} C_{6}$ axis intersect at points $D, E, F$, and $O$, for example, $C_{1} C_{10}$ and $C_{11} C_{2}$ intersect at $D$, so their intersection point lies on the axis. Furthermore, $C_{12} C_{5}$ and $C_{2} C_{10}$ both cut off segments equal to the side length of $S_{12}$ from $C_{1} C_{8}$, because $\angle C_{5} C_{12} C_{1} = \angle C_{12} C_{1} C_{8} = 60^{\circ}$ and $\angle C_{8} C_{1} C_{2} = 90^{\circ}$, $\angle C_{1} C_{2} C_{10} = 45^{\circ}$, thus the intersection point forms an equilateral triangle with $C_{12}$ and $C_{1}$, and an isosceles right triangle with $C_{1}$ and $C_{2}$. Therefore, 3 diagonals intersect at points $D, E, F, O, G$, 5 points in total. $G_{2}$ is the reflection of $G$ across the (unmarked) $C_{1} O$ axis.

Every other intersection point is formed by only 2 diagonals. For example, we will show that $C_{1} E = C_{1} C_{8}$ is intersected by $C_{12} C_{3}$ and $C_{2} C_{11}$ at different points. Indeed, the latter two diagonals are reflections of each other across $C_{1} O$, i.e., $C_{1} C_{7}$, so they intersect at a point $D_{1}$ different from $C_{1}$, which is different from $C_{1} C_{8}$. This can be similarly proven for other seemingly close point triples. The number of intersection points formed by only 2 drawn diagonals is 1 on $O C_{12}$, 4 on $O C_{1}$, and 11 in the triangle $O C_{1} C_{12}$, so $N$ has $5 + 1 + 4 + 22 = 32$ intersection points. In $S_{12}$, there are $6 \cdot 32 + 1 = 193$ intersection points.

The axis $O C_{1}$ passes through 6 sub-regions of $N$, and the triangle $O C_{1} C_{12}$ contains 19 more sub-regions, so the number of sub-regions in $N$ and $S_{12}$ is $6 + 2 \cdot 19 = 44$, and $6 \cdot 44 = 264$.

Note. The number of intersection points formed by only 2 diagonals can also be derived as follows: On each drawn diagonal that can be rotated or reflected to $C_{12} C_{2}, C_{12} C_{3}, C_{12} C_{4}, C_{12} C_{5}$, and $C_{12} C_{6}$, respectively, 4, 12, 13, 18, and 16 other diagonals intersect - for example, on the last one, 2 diagonals start from each of $C_{1}, C_{3}, C_{5}$, and 5 diagonals start from each of $C_{2}$ and $C_{4}$. The number of such diagonals is 6, 12, 6, 12, and 3, respectively, so the number of intersections is $24 + 144 + 78 + 216 + 48 = 510$. In this count, the 3-way intersection points are counted 6 times, and the 2-way intersection points are counted 2 times, so the number of 2-way intersection points is

$$
[510 - 6(1 + 6 \cdot 5)] / 2 = 162
$$

and the total number of intersection points is $6(1 + 6 \cdot 5) + 162 = 193$.

Morvai István (Budapest, Berzsenyi D. g. I. o. t.) Horváth László (Hódmezővásárhely, Ált. Iskola, 7. o. t.)