) \cap B(O,r)  \right) = 1 ~.$$

-   For any $k\in \{1,\ldots,5\}$ and for all integers $n$ such that $0\leq n\leq (R-r_k-1)/2\varepsilon$: $$\hat{N}\left(B\big((r_k+1+2n\varepsilon) e^{\i (2k\pi/5\pm2\varepsilon)},\varepsilon\big) \cap V_{\varepsilon}(r_{1},\ldots,r_{5}) \right)=1 ~.$$

-   The previous points are the only ones of $\hat{N}$.

It is clear that the event $\hat{N}\in B_{\varepsilon}(r_{1},\ldots,r_{5})$ occurs with positive probability, for all $\varepsilon>0$. Roughly speaking, the points of $\hat{N}$ introduced in $(\clubsuit)$ form a chain from $re^{\i 2k\pi/5}$ to $(r_{k}-1)e^{\i 2k\pi/5}$, for any index $k$ such that $r_{k}>r$. See Figure \[fig:5arbres\].

![\[fig:5arbres\] [*RST of the PPP $N$ satisfying both events $A_{\varepsilon}(r_{1},\ldots,r_{5})$ and $B_{\varepsilon}(r_{1},\ldots,r_{5})$. Beware the fact that, in order not to overload the figure, the condition $(\diamondsuit)$ of $B_{\varepsilon}(r_{1},\ldots,r_{5})$ has not been represented. The two balls are centered at $O$ with radii $r=\min_{k\in \{1,\dots,5\}} r_{k}$ and $R=\max_{k\in \{1,\dots,5\}} r_{k}+1$. The $X_{k}$’s are represented by big gray squares while the other points of $N$ by small black circles.*]{}](5arbres.eps){width="11cm" height="11cm"}

On Figure \[fig:5arbres\], imagine that $R=r_{4}+1$ is much larger than $r=r_{5}$ (indeed, we have no control on the $r_{k}$’s). Henceforth, the semi-infinite path $\gamma_{5}$ could prefer to branch on the points of $\hat{N}$ introduced in $(\clubsuit)$ and with direction $8\pi/5$ rather than on $X_{5}$. To prevent this situation from occurring, we contain each path $\gamma_k$ in the cone $C_{2k\pi/5,3\varepsilon,r_{k}}$ thanks to the points of $\hat{N}$ introduced in $(\diamondsuit)$. These points form “landing runways” for the $\gamma_{k}$’s (they may also change slightly the $\gamma_{k}$’s).\
Let us denote by $A_{\varepsilon}$ and $B_{\varepsilon}$ the events $A_{\varepsilon}(r_{1},\ldots,r_{5})$ and $B_{\varepsilon}(r_{1},\ldots,r_{5})$. Then, $$\{ \widetilde{N