# Edit Edit an Image Page: The Party Problem (Ramsey's Theorem)

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{{!}}- {{!}} [[Image:Six2a.png|left]] {{!}} rowspan=2 {{!}} Remember that we are trying to find a configuration that does not produce any monochromatic triangles. Look at the triangle that would be formed by vertices A, B, and C. Since edges AB and AC are blue, we would have to make BC red to prevent triangle ABC from becoming monochromatic. Likewise, in triangle ACD, since edges AC and AD are blue, edge CD must be red. {{!}}- {{!}} colspan=2 {{!}}
{{!}}- {{!}} [[Image:Six3a.png|left]] {{!}} rowspan=2 {{!}} Now, look at the triangle that would be formed by vertices A, B, and D. Since edges AB and AD are blue, edge BD must be red to avoid a monochromatic triangle. Even though we avoided creating a blue monochromatic triangle, we are eventually forced to form a red monochromatic triangle that has the vertices B, C, and D. {{!}}- {{!}} colspan=2 {{!}}
{{!}}- {{!}} [[Image:VertexAni1a.gif|left]] {{!}} rowspan=2 {{!}} This holds true even if we change our initial assumption that all the edges emanating from A are blue. If edge AE were red, edge AF were red, or both edges AE and AF were red, our end results would not change. This is because these are trivial changes to the configuration. We didn't use or consider edges AE and AF when we formed the monochromatic triangle BCD, so changing their colors has no effect on triangle BCD. {{!}}- {{!}} colspan=2 {{!}}
{{!}}- {{!}} [[Image:VertexAni2a.gif|left]] {{!}} rowspan=2 {{!}} Now what about nontrivial changes to the configuration? Say we change edge AD from blue to red. If AD isn't blue, we can no longer put a constraint on edge CD—it can be either red ''or'' blue. It is no longer guaranteed that triangle BCD is monochromatic. {{!}}- {{!}} colspan=2 {{!}}
{{!}}- {{!}} [[Image:Six5a.png|left]] {{!}} rowspan=2 {{!}} Let's look at this new configuration again. Edges AC and AF are blue, so edge CF must be red to prevent triangle ACF from becoming a blue monochromatic triangle. Edges AB and AF are blue, so edge BF must be red to prevent triangle ABF from becoming blue and monochromatic. Sound familiar? We are, once again, forced to form a red monochromatic triangle (triangle BCF). {{!}}- {{!}} colspan=2 {{!}}
{{!}}- {{!}} [[Image:Six6.png|left]] {{!}} rowspan=2 {{!}} Even if we make a nontrivial change to the configuration and change edge AC from blue to red, we are still forced to form the red monochromatic triangle BEF. {{!}}}