Triangle Parenting

Relevant wiki: Similar Triangles - Problem Solving - Medium

Let $a$ , $b$ , $c$ be complex numbers corresponding to the vertices of one parent triangle $\bigtriangleup ABC$ in the complex plain and $\omega = \frac{a-b}{c-b}$ be a spiral simmetry. A triangle mustpresent the same spiral simmetry (or the conjugate one) between some of its sides in order to be similar to $\bigtriangleup ABC$ .

Let's say $\bigtriangleup ABC$ mates with triangle $\bigtriangleup XYZ$ and their child is similar to, for example, $\bigtriangleup ABC$ itself. Then: $\frac{\frac{(a+x)}{2} - \frac{(b+y)}{2}}{\frac{(c+z)}{2} - \frac{(b+y)}{2}} = \omega$ $\Leftrightarrow\; (a-b) + (x-y) = \omega (c-b) + \omega (z-y) \;\Leftrightarrow\; \frac{x-y}{z-y} = \omega$ If nothing strange happens (and we can make it not happen), than $\bigtriangleup XYZ$ and $\bigtriangleup ABC$ must be directly similar to each other and $X$ , $Y$ , $Z$ must be connected to $A$ , $B$ , $C$ in this order!

Thus $\left(\bigtriangleup_x,\bigtriangleup_y\right)$ have a child similar to them in any possible case iff they are directly similar and corresponding vertices have are named with the same letter. We can permute the vertices' name in $3! = 6$ possible ways.

With $12$ triangles only we could have half triangles clockwise similar and all differenlty labeled and the other half counterclokwise similar and equally differently labeled.

But if we have $13$ triangles at least $7$ must be directly similar by the pigeonhole principle!

Having $7$ of them, at least $2$ must have corresponding vertex equally named, again, by the pigeonhole principle!

Hence, we need $\boxed{n\geq 13}$ .