Five vertices of a regular $10$-gon are painted red and five blue. Prove that there will always be two congruent monochromatic triangles.
Please tell me if my proof is acceptable.
I don't know how to draw in LaTeX .... But. What I did was draw a circle and then inscribe a $10$-gon in it. I start naming the vertices of the polygon in clockwise order $A,B, \dots I, J$.
Now, consider the five diametrically opposite pairs $(A,F), (B,G) , (C,H), (D,I), (E,J)$.
If there are no such monochromatic pairs, then any monochromatic triangle is congruent to the triangle drawn from the diametrically opposite points of each of its vertices.
If there is one monochromatic pair of red, there has to be one monochromatic pair of blue too. However! we still have three pairs that are differently colored. The two monochromatic triangles from these pairs will be congruent.
If there are two monochromatic red pairs, there are two monochromatic blue pairs. But, there has to be at least one pair of differently colored vertices. We select this differently colored pair $(R,B)$, and We get two congruent triangles by connecting the red vertex, to a red pair and the blue vertex to a blue pair.
There is a counter example to this. If $A, C, D, H, I$ are red, and $B, E, F, G, J$ are blue. Then, connecting the differently colored pair to a pair will not give congruent triangles.