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Given any $10$ people in a room, prove that among them there are either $3$ that know each other or $4$ that do not know each other. (Assume that if A knows B, then B knows A.)

Could the pigeonhole principle be used for this? If so, how? I have a feeling it can be used but im not sure where to apply it just yet.

user1176409
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Bippulli
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3 Answers3

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Choose one of the present people, say $A$. The rest are split into two groups- those that know $A$ (let's say, group $T$) and those that don't (let's say, group $S$). There are just two possibilities: either $|S| ≥ 6$ or $|T| ≥ 4$ (or otherwise $|S∪T| < 9$ which is a contradiction).

If $|S| ≥ 6$, then there has to be $3$ members of $S$ that don't know each other (a well known result). Together with $A$, they form a group of four mutual strangers.

If $|T| ≥ 4$, then either they all don't know each other (in which case we are already done), or some two are acquaintances. In the latter case, together with $A$ these two form a triple of mutual acquaintances.

This completes the proof.

Sayan Dutta
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    You have swapped know and don't know, though it is not important. The extension to $9$ in a room says something similar, but for every point you need $|S|\ge 6$ (done as in the $10$ case), or $|T|\ge 4$ (done as in the $10$ case), or $|S|=5$ and $|T|=3$. But the last of these is impossible as it would lead to the number of pairs who know each other being $\frac{27}{2}$ and the number of pairs who do not know each other being \frac{45}{2}$, neither of which are integers – Henry Oct 06 '21 at 21:59
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    In fact you can prove $1$ monochrome triangle for $6$ ,$2$ triangles for $7$ , etc. – Roddy MacPhee Oct 06 '21 at 23:39
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    @Henry Thanks for pointing that out. Please check whether it's okay now. – Sayan Dutta Oct 07 '21 at 14:54
  • @RoddyMacPhee Actually there are (at least) $2$ monochromatic triangles for $6$ vertices, $4$ monochromatic triangles for $7$ vertices. – bof Jul 24 '23 at 08:48
  • @bof yeah technically any even number n-4 – Roddy MacPhee Jul 26 '23 at 23:36
  • @RoddyMacPhee Huh? The minimum possible number of monochromatic triangles in a $2$-coloring of the edges of the complete grapg $K_n$ is given by Goodman's formula: $$\binom n3-\left\lfloor\frac n2\left\lfloor\left(\frac{n-1}2\right)^2\right\rfloor\right\rfloor$$ – bof Jul 27 '23 at 03:25
  • Okay fine I'm not formula based. Once you show two for n equal to 6 you can easily show n-4 for even values is possible. Was my point. – Roddy MacPhee Jul 27 '23 at 18:10
  • @RoddyMacPhee Oh, never mind Goodman's formula. Once you show that in a $2$ -coloring of $K_6$ there are $2$ monochromatic triangles, i.e. 10% of the $20$ triangles, it follows that in a $2$-coloring of $K_n$ where $n\ge6$, at least 10% of the triangles are monochromatic, i.e. at least $\binom n3/10$ of them. So for $n=10$ there are at least $12$ monochromatic triangles. (Actually, by Goodman's theorem, there must be at least $20$.) – bof Jul 28 '23 at 06:13
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It's very simple, let A be an arbitrary person among the 10 people, and denote the other 9 people from $A_1$ to $A_9$.

Case I: A knows at least 4 of $A_i$'s. Then assume A knows $A_1,A_2,\cdots, A_5$.

if $A_i, A_j$ knows each other, then we have a triangle $A, A_i, A_j$. Otherwise, $A_1,\cdots, A_5$ constitutes a set of 5 people that don't know each other

Case II: A doesn't know at least 6 of $A_i$'s. Then consider 6 people $A_1, A_2, A_3,A_4,A_5,A_6$.

Then apply the following lemma shall yield the easy solution:

Lemma There exist three people who mutually know/or don't know each other among 6 random people.

There's a stronger lemma I think you also might want to consider, that is to prove that there are two triangles of people who mutually know/don't know each other.

Nikola Tolzsek
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Answered the same at a duplicate, but as it was closed I will transfer it here. Basically we need to show that: $R(4,3) \leq 10 $ but:

$$ R(4,3) \leq R(4,2)+ R(3,3) = 4+ 6 =10 $$

And it is pretty straightforward to show that $R(4,2)=4$ and that $R(3,3)=6$

SotArmen
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