So I think I have the answer to this problem, but there's something about it that's bothering me:
Suppose we choose a fixed point with $13$ edges coming out of it.
There must be at least
$a)$ $9$ red edges (friends)
OR
$b)$ $5$ blue edges (strangers)
because $8 + 4 < 13$.
In case $b)$, among the $5$ edges if any are blue then we have blue triangle with our fixed point, and so we are done. Otherwise, they are all red and we have a red-pentagon, also as desired.
In case $a)$, among $9$ people there are $3$ mutual strangers or $4$ mutual friends. If the former, we are done. If the latter, since the $4$ friends are red edges to our fixed point then we have $5$ mutual friends including our fixed point, again as desired.
Namely, for case $a)$, do I need to include the case if among $9$ people there are $4$ mutual STRANGERS or $3$ mutual FRIENDS? If so, how? If not, well why not?
This has been bothering me for quite some time and I would appreciate an answer!
Thanks!
Your solution to the problem is good. Ofcourse this relies on the fact that among 9 people there allways are 3 mutual strangers or 4 mutual friends. This is a known bound so you can use it, but if this is some kind of hand in exercise then you may have to do that step too.
Thus either you use the known proposition or you prove it. In both cases you do not have to consider 4 mutual strangers and 3 mutual friends, since you still know that there are 3 mutual strangers and 4 mutual friends.