Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?

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Through a non-central point inside a regular dodecagon, what is the maximum number of different diagonals that can be drawn?

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Through a non-central point inside a regular dodecagon,  
what is the maximum number of different diagonals that can be drawn?
Let the regular dodecagon be $A_{1} A_{2} \cdots A_{12}$.

First, prove that the 4 diagonals $A_{1} A_{5}, A_{2} A_{6}, A_{3} A_{8}, A_{4} A_{11}$ intersect at a non-central point inside the regular dodecagon.

Since $\overparen{A_{5} A_{8}}=\overparen{A_{8} A_{11}}$, $\overparen{A_{11} A_{1}}=\overparen{A_{1} A_{3}}$, and $\overparen{A_{3} A_{4}}=\overparen{A_{4} A_{5}}$, the lines $A_{3} A_{8}$, $A_{5} A_{1}$, and $A_{11} A_{1}$ are the angle bisectors of $\triangle A_{3} A_{5} A_{11}$. Therefore, $A_{1} A_{5}$ passes through the intersection point $K$ of the diagonals $A_{3} A_{8}$ and $A_{4} A_{11}$. Similarly, the lines $A_{2} A_{6}$, $A_{4} A_{11}$, and $A_{8} A_{3}$ are the angle bisectors of $\triangle A_{2} A_{4} A_{8}$, so the diagonal $A_{2} A_{6}$ also passes through point $K$. Thus, these 4 diagonals intersect at point $K$. Point $K$ lies on the chord $A_{1} A_{5}$, and $K$ is different from the center $O$.

Next, prove that in a regular dodecagon, there do not exist 5 diagonals intersecting at a non-central point inside the dodecagon.

Use proof by contradiction. Assume that such 5 diagonals exist. Then, the 10 endpoints of these 5 diagonals are all distinct, and any one of these diagonals divides the circumcircle of the dodecagon into two arcs, each containing 4 endpoints of the other 4 diagonals. This implies that if these 10 endpoints are given, then the 5 diagonals are also determined.

By the symmetry of the vertices of the regular dodecagon, without loss of generality, assume that among the 12 vertices, excluding $A_{1}$ and one of $A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$, these 10 endpoints are given. Now, consider the cases for this one point.
(1) If this point is $A_{2}$, since $\overparen{A_{1} A_{7}}=\overparen{A_{9} A_{12}}$, the intersection point $M$ of the diagonals $A_{4} A_{9}$ and $A_{7} A_{12}$ lies on the line through the center $O$ and point $A_{8}$. This can be seen by noting that $\triangle M A_{4} A_{7}$ is congruent to $\triangle M A_{12} A_{9}$, and $\triangle O M A_{9}$ is congruent to $\triangle O M A_{7}$. Thus, the intersection point $M$ does not lie on the other diagonal $A_{3} A_{\varepsilon}$, leading to a contradiction.
(2) If this point is $A_{3}$, since $\overparen{A_{4} A_{5}}=\overparen{A_{9} A_{10}}$, the intersection point $M$ of the diagonals $A_{4} A_{9}$ and $A_{5} A_{10}$ lies on the line through the center $O$ and point $A_{7}$. The point $M$ does not lie on the other diagonal $A_{2} A_{0}$, also leading to a contradiction.
(3) If this point is one of $A_{6}, A_{5}, A_{5}, A_{7}$, then the diagonals $A_{2} A_{8}$ and $A_{3} A_{9}$ are diameters and must intersect at the center $O$, leading to a contradiction.
In conclusion, the answer is 4.