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  "body": "\n## Proof\n\nWe look at the edges at one vertex and use the pigeonhole principle.\n\n1. Colour each edge of the [[complete graph|03WKB2H474]] $K_6$ red or blue. Fix a vertex $v$.\n2. The vertex $v$ is adjacent to the other $5$ vertices, so it lies on $5$ edges.\n3. Suppose that at most $2$ of these edges were red and at most $2$ were blue. Then there would be at most $4$ edges, a contradiction. This is the [[pigeonhole principle|0HGVCWQGY3]] for $5$ edges and $2$ colours.\n4. So at least $3$ of these edges have one colour. By symmetry of the two colours, we may call this colour blue.\n5. Let $a, b, c$ be three distinct vertices such that $va$, $vb$ and $vc$ are blue.\n6. Case 1: one of the edges $ab$, $bc$, $ca$ is blue, say $ab$. Then the edges $va$, $vb$ and $ab$ between $v, a, b$ are all blue.\n7. Case 2: none of $ab$, $bc$, $ca$ is blue. Then all three are red, and the three edges between $a, b, c$ are all red.\n8. In both cases there are three vertices such that the three edges between them have the same colour.\n\n## Idea\n\nSome vertex sees three others along edges of the same colour, say blue. Among these three, a single blue edge closes a blue triangle with the vertex. If there is no blue edge among them, they form a red triangle.\n",
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