Problem 1142

Proof: We prove the strengthened proposition by induction for $n \geq 5$: If the edges of a convex $n$-gon are colored with three colors $a, b, c$, and each color is used at least once, then a triangular triangulation satisfying the requirements can be made. $\qquad$

When $n=5$, if the number of edges of the three colors is $1,1,3$, by symmetry, we only need to consider the following two cases, each of which can be triangulated as shown in the figure.

If the number of edges of the three colors is $1,2,2$, by symmetry, we only need to consider the following three cases, each of which can be triangulated as shown in the figure.
Assume the conclusion holds for $n (n \geq 5)$, and consider the case for $n+1$, where the convex $(n+1)$-gon is denoted as $A_{1} A_{2} \cdots A_{n+1}$.
Case 1: There are two colors of edges, each with only one edge. Without loss of generality, assume the $a$ and $b$ colored edges each have only one edge. Since $n+1 \geq 6$, there exist two consecutive edges that are both $c$ colored, without loss of generality, let these be $A_{n} A_{n+1}$ and $A_{n+1} A_{1}$. Draw the diagonal $A_{1} A_{n}$ and color $A_{1} A_{n}$ with $c$ color, then the triangle $A_{n} A_{n+1} A_{1}$ has all three sides of the same color. At this point, the convex $n$-gon $A_{1} A_{2} \cdots A_{n}$ has at least one edge of each color, and by the induction hypothesis, it can be triangulated to meet the requirements. $\qquad$
Case 2: One color of edge has only one edge, and the other colors have at least two edges each. Without loss of generality, assume the $a$ colored edge has only one edge. We can choose two adjacent edges that are not $a$ colored, without loss of generality, let these be $A_{n} A_{n+1}$ and $A_{n+1} A_{1}$. Draw the diagonal $A_{1} A_{n}$, then $A_{1} A_{n}$ has a unique coloring such that the triangle $A_{n} A_{n+1} A_{1}$ has all three sides of the same color or all different colors. At this point, the convex $n$-gon $A_{1} A_{2} \cdots A_{n}$ has at least one edge of each color, and by the induction hypothesis, it can be triangulated to meet the requirements. $\qquad$
Case 3: Each color of edge has at least two edges. Draw the diagonal $A_{1} A_{n}$, then $A_{1} A_{n}$ has a unique coloring such that the triangle $A_{n} A_{n+1} A_{1}$ has all three sides of the same color or all different colors. At this point, the convex $n$-gon $A_{1} A_{2} \cdots A_{n}$ has at least one edge of each color, and by the induction hypothesis, it can be triangulated to meet the requirements. Combining the above three cases, we see that the conclusion also holds for $n+1$.
By mathematical induction, the conclusion is proven. 50 points