ParthKohli
  • ParthKohli
@ganeshie8
Mathematics
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chestercat
  • chestercat
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ParthKohli
  • ParthKohli
Show that if the 21 edges of a complete 7-gon (all vertices joined to one another) is coloured red and blue, there exist at least three monochromatic triangles.
ParthKohli
  • ParthKohli
Obviously a pigeonhole thing, but how do I approach it?
ganeshie8
  • ganeshie8
|dw:1446569391614:dw|

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ParthKohli
  • ParthKohli
Question: can an edge be a part of two triangles which satisfy this? I think yes.
ganeshie8
  • ganeshie8
Yes, each of the \(21\) edges is formed by choosing \(2\) vertices from the available \(7\) vertices : \(\binom{7}{2}\)
ParthKohli
  • ParthKohli
Yes, of course. The simplest answer to my question is to just consider all edges coloured red.
ganeshie8
  • ganeshie8
I need to go for dinner, but here is what I have in mind : |dw:1446570150329:dw|
ganeshie8
  • ganeshie8
1) From the vertex \(A\), there are \(6\) edges shooting out. 2) By pigeonhole principle at least \(\dfrac{6}{2}=3\) of them will have same color (call it red for definiteness). 3) ...
ganeshie8
  • ganeshie8
that doesn't prove anything, yet.. we need to show that at least 3 "triangles" will have all edges same color
ParthKohli
  • ParthKohli
yes, exactly.
ParthKohli
  • ParthKohli
how is a triangle formed? like is the condition that two of the sides have to be originating from the same vertex?
ParthKohli
  • ParthKohli
|dw:1446571044672:dw|
ParthKohli
  • ParthKohli
Is OBC a triangle? I mean it is of course, but does the problem count it as one?
ParthKohli
  • ParthKohli
or is it formed strictly by choosing three out of the seven vertices?\[\binom{7}3 = 35\]
ParthKohli
  • ParthKohli
Just found the same problem restated in graph theory, and wow, actually we can even prove the existence of four monochromatic triangles.
ParthKohli
  • ParthKohli
How about this one? For real \(x\), show that x, 2x, ..., (n-1)x contains at least one number that differs from an integer by at most \(\frac{1}n\).

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