Monochromatic congruent triangles on a 10-gon

Question

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.

Answer 1

It's fine until your last step. There you have a pair of differently coloured vertices, and a blue pair, and a red pair. It is not in general true that doing what you did gives you two congruent triangles.

Here's the simplest way I found to continue from your idea. If there are two differently coloured pairs (of opposite vertices), they form a rectangle with one side red and the other side blue, and no other pair can be differently coloured, but there are an odd number of pairs left so it's impossible. Therefore there is exactly one differently coloured pair. Now it's easy to see that it's impossible as there is essentially one forced way that ends in a problem, namely if $A_1$ is red and $A_6$ is blue then by symmetry of colours we can assume that $A_2$ is red, which means $A_7$ is red, and so $A_0,A_5$ cannot be blue, but then all the remaining $4$ vertices must be blue, and you can now easily find congruent triangles of different colour.

Answer 2

Let me call a pair of diametrically opposite same-colored vertices a monochromatic diameter, and let me call a pair of diametrically opposite different-colored vertices a bichromatic diameter. Sounds ugly, but I have to call them something.

You noted correctly that, if you have $5$ bichromatic diameters, they you get a pair of congruent monochromatic triangles of different colors. But you don't need $5$ bichromatic diameters for that, all you need is $3$ of them! So, to finish the proof, you only have to consider colorings where the number of bichromatic diameters is less than $3.$

But the number of bichromatic diameters has to be an odd number! This is becaus you have equal numbers of red and blue vertices, so the monochromatic diameters have to split evenly into red-red and blue-blue. Take away an even number of monochromatic diameters from $5,$ you're left with an odd number of bichromatic diameters. So the number of bichromatic diameters is an odd number less that $3,$ i.e., there is exactly one bichromatic diameter.

Without loss of generality we can assume that $AF$ is the one and only bichromatic diameter; say $A$ is red and $F$ is blue. Now we just need to know the colors of the vertices $B,C,D,E;$ the diametrically opposite vertices will have the same colors. Since two of them are red and two blue, there are only $\binom42=6$ cases. That's small enough to finish off with a little brute force.

Namely, the $6$ cases for the set of red vertices are $ABCGH,\ ABDGI,\ ABEGJ,\ ACDHI,\ ACEHJ,\ ADEIJ.$
You just have to show that in each of those $6$ colorings there are two congruent monochromatic triangles.

This should work, but no doubt there's a more elegant proof. It looks like someone has posted one while I was typing this.