11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles triangles formed by the 10 points $A, B, C, D, E, F, G, H, I, J$ and the line segments in Figure 2 be denoted as $M$.
(1) Determine the number of elements in $M$;
(2) If each of the 10 points is arbitrarily colored either red or blue, does there necessarily exist an isosceles triangle in $M$ whose three vertices are the same color?
(3) If $n$ of the 10 points are arbitrarily colored red such that there necessarily exists an isosceles triangle in $M$ whose three vertices are all red, find the minimum value of $n$.
11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the same color, and the triangle formed by these three points belongs to $M$. Therefore, there must exist an isosceles triangle in $M$ whose three vertices are of the same color.
(3) If $n=5$, then coloring $F, G, H, I, J$ red, there will be no triangle in $M$ whose vertices are all red.
If $n=6$, when there are at least three red points among $A, B, C, D, E$, there must exist a triangle in $M$ whose vertices are all red; when there are fewer than three red points among $A, B, C, D, E$, there are at least four red points among $F, G, H, I, J$.
If there are exactly four red points among $F, G, H, I, J$, assume $F, G, H, J$ are red points, then there are at least two red points among $A, B, C, D, E$. If one of $A, B, E$ is a red point, assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red; otherwise, $C, D$ are both red, thus, $\triangle CDH$ is a triangle in $M$ whose vertices are all red.
If $F, G, H, I, J$ are all red, then there is at least one red point among $A, B, C, D, E$. Assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red. Therefore, the minimum value of $n$ is 6.
(Li Jianquan provided)