Nodes and triangles

Pick any vertex $z$ . It connects to five other vertices, and so at least three of the edges coming from it must be of the same colour. Suppose, for definiteness, that three vertices $z_1,z_2,z_3$ are connected to $z$ with red edges. If any two of these vertices, say $z_i$ and $z_j$ , are connected by a red edge, then $z,z_i,z_j$ is a red triangle. Thus no two of these vertices are connected by a red edge, and hence $z_1,z_2,z_3$ is a blue triangle. Thus at least one monochromatic triangle exists.

However we have found them, suppose that the vertices $w_1,w_2,w_3$ are connected by a monochromatic triangle. For definiteness, suppose that this triangle is red. Suppose that there are no other red triangles. If $u_1,u_2,u_3$ are the other three vertices, then two of them must be connected by a blue edge (otherwise we would get a new second red triangle). Suppose that the edge $u_1u_2$ is blue. If any two of the edges connecting $u_1$ to $v_1,v_2,v_3$ were red, say $u_1v_i$ and $u_1v_j$ , then $u_1,v_i,v_j$ would be the vertices of a red triangle. Thus at least two of the edges $u_1v_1,u_1v_2,u_1v_3$ must be blue. Similarly, at least two of the vertices $u_2v_1,u_2v_2,u_2v_3$ must be blue. Consequently there must be some $1 \le k \le 3$ such that $u_1v_k,u_2v_k$ are both blue, and hence $u_1,u_2,v_k$ are the vertices of a blue triangle. Thus we have found a blue triangle.

Thus there are always at least two monochromatic triangles.

Consider a node $A$ . WLOG, assume it connects to $\ge3$ blue edges. In the $\ge3$ nodes that $A$ connects to: if any connect to each other with a blue edge, then we have formed a triangle; if all of them connect to each other with red edges, then we have again formed a triangle. Therefore, there exists $\ge1$ triangle.

WLOG, start with a blue triangle:
To avoid generating a new triangle (using the method described previously) from a disconnected node, that node must use red edges to connect to the $3$ nodes currently making a triangle, and use blue edges to connect to the remaining nodes (see comments). Hence, we get this:
and we find that we have made $2$ triangles.

The answer of this question is yes .

To show why, let's try to connect all of the six vertices without creating two triangles. The TED-Ed video that answers a similar problem explains why you will always have at least one complete triangle with six vertices. So WLOG, let's create that triangle as follows:

We're going to call the three nodes on the left that already form a triangle the left nodes and the three on the right that currently have no edges the right nodes. Each of the right nodes must connect to all of the left nodes. If a specific right node connects to the left nodes with 2 or 3 blue lines we create a second blue triangle, which is what we want to avoid. So let's connect each of the right nodes to all of the left nodes with only red edges. That way we don't created any new triangles. Plus, we just gave ourselves some extra freedom. For each of the three right nodes we may freely change one of its edges to blue later on without creating any new triangles. We now have the following:

We only need to connect the three right nodes with each other. If we make all of those edges blue we just created a second triangle, so we can't do that. If we make two or three of the edges red we create very many red triangles, so we also can't do that. There's only one thing to do: create two blue edges and one red edge. WLOG, let's connect the top-right and the bottom-right nodes with that red edge:

This creates three red triangles. But no worry, earlier we gave ourselves the freedom to change 3 red edges to blue, so let's use them to get rid of those triangles. We can use the one for the top-right node to destroy one red triangle, the one from the bottom-right node to destroy the second and finally we use the one of the right node to destroy the third tri.... wait. None of the red triangles we have uses the right node, so changing one of its edges to blue won't help.

That means we have to accept that no matter how you color the edges, there will always be at least two complete triangles, so the answer is yes . $\square$

$C_6^2$ = 15 edges, $C_6^3$ = 20 triangles.

Suppose vertex $v_i$ is connected by $r_i(0 \leq i \leq 5)$ red edges and $(5 - r_i)$ blue edges. Because $r_i, 5-r_i \leq 6$ , the number of heterochromatic angles (angles have two different colored sides) $\leq 6*6 = 36$ . And because one heterochromatic triangle has two heterochromatic angles, the number of those triangles $\leq 18$ . Thus the number of complete triangles $\geq 20 - 18 = 2$ .

There are at least 2 triangles. We can prove it by writing an algorithm guaranteed to "find" the other triangle based on the first triangle.

We already know there's at least one triangle. Label the nodes of that triangle A , B and C . We also assume WLOG (without loss of generality) that the triangle is red.

Consider the triangle formed by the remaining nodes. It is either a red triangle, a blue triangle, or neither. If it's a red or blue triangle, we have our 2nd triangle. Otherwise, there must be at least 1 blue line and 1 red line in the triangle. Look at the blue line (if there are 2 blue lines just pick either one), and label the nodes at each end D and E .

Consider lines AD , BD and CD . If any 2 of them are red, then we have our 2nd red triangle. Otherwise, at least 2 of them are blue.

Assume WLOG that AD and BD are blue. If AE is blue, then we have a blue triangle in ADE . Similarly, if BE is blue, then we also have a blue triangle in BDE . Otherwise, AE and BE have to be red. But then we would have a 2nd red triangle in ABE .

So there will be a 2nd red triangle or a blue triangle no matter what.

To simplify, the hexagon shape can only be filled with the following:

Since can be the colour reflection, we can remove the duplication as such we only need to do 6 blue, 5 blue 1 red, 4 blue 2 red and 3 blue 3 red shapes. Rotate symmetrical shapes are taken out as well as they form the same triangles.

Before drawing we have to obey these 2 rules: Rule 1: The internal lines are drawn opposite so that we do not make a coloured triangle. Rule 2: Purple lines are drawn to act as a wild colour which can replace red or blue lines

For 6 blue as you can see if any red lines change to blue it becomes a blue triangle, the red lines form at least 2 triangles.

For 5 blue 1 red, whereby if it is red or blue it forms a red triangle or blue triangle respectively. Therefore form 2 coloured triangles.

For 4 blue 2 red, the combination is

or

or

last but not least For 3 blue 3 red, the combination is

or

or whereby we reverse engineered and draw a few blue lines

Why are there black edges in the diagram? Am I missing something?