Denote by $l $ the triangle's side. Suppose $A $ and $P $ have a common element. Let $x $ be its length. Now, we are looking at $A $: Every surface in which $ABC $ is divided has the area $\frac {l^2 \sqrt 3}{4n}$, so, if we denote by $k $ the counter of the parallel of length $x $(from $A $ to $BC $), we will have that $x=\sqrt {\frac {l^2 k}{n}} $. Now we are looking at $P $: Denote by $x_1$, $x_2$, $... $, $x_n $ the length of the parallels from $A $ to $BC $, where $x_n$ equals $l $. Because it does not hurt, let $x_0$$=0$. Every perimeter will be equal with $3x_1$. So we have: $3x_1 = x_k + x_{k-1}+2 (x_k -x_{k-1})$. Then$ 3 (x_k - \frac {3x_1}{2})=x_{k-1} -\frac {3x_1}{2}$. So, using $x_n =l $, we have that $x_1=\frac {2*3^{n-1}l}{3^n -1} $. Consequently, $x_k=\frac {3^{n-k} (3^k -1)l}{3^n -1} $. By our assumption, there exist $i $ such that $x_i =x $, or $\frac {k}{n}= \frac {3^{2n-2i} (3^i -1)^2}{(3^n -1)^2}.$ Now it follows that $3^{2n-2i} $ divides $k $ and $3^i $ divides $n-p $, where $p $ is natural number with $k=3^{2n-2i} p $. So $3^i \leq n$ and $3^{2n-2i} \leq n $. From here, we obtain that $9^n \leq n^3$ which is false by induction. Done