XIII OBM

Interpreting the points $P_k$ as vectors, we have $P_{n+4}=\frac{P_n+P_{n+1}}{2}$, which is a linear homogeneous recursion. Its characteristic polynomial is $2x^4-x-1=(x-1)(2x^3+2x^2+2x+1)$. So let $\alpha ,\beta ,\gamma$ be the roots of $f(x)=2x^3+2x^2+2x+1$, so that $P_n=Q_0+Q_1{\alpha }^n+Q_2{\beta }^n+Q_3{\gamma }^n$, $Q_0,Q_1,Q_2,Q_3$ being constant vectors.

We have ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )={\left(-\frac{2}{2}\right)}^2-2\cdot \frac{2}{2}=-1<0$, so $f$ has one real root and two conjugate complex roots. Suppose $\alpha$ is real. Since $f(-1)=-1<0$ and $f\left(-\frac{1}{2}\right)=\frac{1}{4}>0$, $-1<\alpha <-\frac{1}{2}$ and $|\beta |=|\gamma |=\sqrt{|-\frac{1}{2}|}<1$, so all roots of $f$ have modulus smaller than 1 and so $P_n$ tends to $Q_0$ as $n$ goes to infinity. This means that $Q_n$ tends to $Q_0$ and thus the intersection of all $Q_n$ is $Q_0$. To find $Q_0$, notice that

$$|\begin{array}{l}{P}_0=Q_0+Q_1+Q_2+Q_3\\ 2P_1=2Q_0+Q_1\cdot 2\alpha +Q_2\cdot 2\beta +Q_3\cdot 2\gamma \\ 2P_2=2Q_0+Q_1\cdot 2{\alpha }^2+Q_2\cdot 2{\beta }^2+Q_3\cdot 2{\gamma }^2\\ 2P_3=2Q_0+Q_1\cdot 2{\alpha }^3+Q_2\cdot 2{\beta }^3+Q_3\cdot 2{\gamma }^3\\ \Rightarrow P_0+2P_1+2P_2+2P_3=7Q_0+f(\alpha )Q_1+f(\beta )Q_2+f(\gamma )Q_3=7Q_0\\ ⟺Q_0=\frac{P_0+2P_1+2P_2+2P_3}{7}=\left(\frac{3}{7},\frac{4}{7}\right)\end{array}$$