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A problem in my book is:

Let the edges of $K_7$ be colored with the colors red and blue. Show that there are at least four subgraphs $K_3$ with all three edges the same color (monochromatic triangles). Also show that equality can occur.

By the theorem on friends and strangers it is clear that 1 monochromatic triangle exists. Deleting a vertex of that triangle and applying the theorem again yields another. Why are two more guaranteed?

As an aside, a result in my book states that the number of monochromatic triangles in a 2-colored $K_n$ is at least $\binom{n}{3}-\lfloor \frac{n}{2}\lfloor (\frac{n-1}{2})^2 \rfloor \rfloor $. I want to demonstrate my solution without applying this result though as it appears later on in the book.

Thank you for your time.

InvisiblePanda
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Shahab
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2 Answers2

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A "biangle" is a triple of vertices $(a,b,c)$ where the edge joining $a$ to $b$ is not the same color as the edge joining $b$ to $c$. We call $b$ the "apex" of the biangle.

A vertex with 3 red and 3 blue edges is the apex of 9 biangles; any other color distribution leads to fewer biangles. Thus, the whole graph has at most 63 biangles.

If a triangle is not monochromatic, it has exactly 2 biangles. So the graph has at most 31 non-monochromatic triangles.

But the graph has 35 triangles, so it has at least 4 monochromatic triangles.

Gerry Myerson
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Gerry Myerson's answer is very good. However, I wanted to offer a different, more simple-minded answer.

To begin with, my answer requires that I show every $K_6$ has two monochromatic triangles. The proof is as follows:

Suppose there is only one monochromatic triangle As you noted, we know that every $K_6$ must have at least one monochromatic triangle. For ease of reference, I call the vertices in this monochromatic triangle $A$, $B$, and $C$; I call the other three vertices $D$, $E$, and $F$. Suppose, without loss of generality, that the triangle formed by $A$, $B$, and $C$ is red. At most one of the edges connecting $D$ to $A$, $B$, and $C$ can be red; otherwise we have must have at least one more red monochromatic triangle. We can reason similarly for $E$ and $F$. Now take $D$ and $E$, and notice that, by the Pigeonhole Principle, there must be at least one vertex of $A$, $B$, and $C$ such that $D$ and $E$ are both connecting by a blue edge to that vertex. If $D$ and $E$ are connected by a blue edge, then we have a monochromatic triangle, so they must be connected by a red edge. We can reason similarly for the edge connecting $D$ and $F$ and the edge connecting $E$ and $F$. But then $D$, $E$, and $F$ must form a red triangle. So we have a contradiction and there must be more than one monochromatic triangle in $K_6$.

Now I turn to the involving $K_7$.

For ease of reference, I refer to the seven vertices in $K_7$ by the letters $A$, $B$, $C$, $D$, $E$, $F$, and $G$. The subgraph of $K_7$ made up of $A$, $B$, $C$, $D$, $E$, and $F$ must have two monochromatic triangles. Without loss of generality, suppose $A$ is a vertex in one of the monochromatic triangles of the subgraph with vertices $A$, $B$, $C$, $D$, $E$, and $F$. Then the subgraph made up of up $B$, $C$, $D$, $E$, $F$, and $G$ has a monochromatic triangle that wasn't in the subgraph made up of $A$, $B$, $C$, $D$, $E$, and $F$ or the subgraph made up of $A$, $B$, $C$, $D$, $E$, and $F$ had at least three monochromatic triangles at least two of which are in the subgraph consisting of $B$, $C$, $D$, $E$, and $F$. Either way, this means that $K_7$ has at least three monochromatic triangles. If $K_7$ has three monochromatic triangles, then by Pigeonhole Principle at least one vertex must be in two monochromatic triangles, which means the subgraph of $K_7$ consisting the six vertices other than this vertex must have at least two monochromatic triangles. So $K_7$ must have at least four monochromatic triangles.