Max Noether theorem application problem I've been trying for a while this problem, thinking about different combinations of lines and the given curves to apply Max Noether theorem but can't get to the result. Can I get a hint or a solution? This is the statement:
Let $o$ be a node of both degree $4$ curves $C$ and $C'$. They intersect in another 12 different points $p_1,...,p_{12}$. Show that any curve of degree $5$ that has a triple point on $o$ and passes through 11 of the $p_i$, also passes through the $12^{th}$ point.
 A: $\newcommand{\local}[1]{\mathcal{O}_{\mathbb{P}^2,{#1}}}$
Let $F,G$ be the quartics associated to $C,C'$ and $H$ be the quintic under study.
$F$ and $G$ intersect in $16$ points counted with multiplicity, so they intersect in every $p_i$ with multiplicity $1$ and in $o$ with multiplicity $4$. (Because of $(F \cdot G)_o \geqslant \mu_o(F) \mu_o(G)$, where $(F \cdot G)_o$ is the intersection multiplicity at $o$ and $\mu_o(F)$, $\mu_o(G)$ are the orders of $F$, $G$ at $o$, we have $(F \cdot G)_o \geqslant 4$).
Now consider the ideals $I=(F_o,G_o)$ and $\mathfrak{m}=(x,y)$ in $k[x,y]_{(x,y)} = \local{o}$ and extend them to $k[[x,y]]$. Then, upto a transform of coordinates, we have 
$$F_o k[[x,y]] = (x y + f_3 + f_4 + \cdots) k[[x,y]]$$ 
and 
$$G_o k[[x,y]] = ((x-y) (x+ a y) + g_3 + g_4 + \cdots) k[[x,y]]$$ 
with a certain $a\in k-\{0,-1\}$ and $f_d$, $g_d$ homogeneous of degree $d$.
So 
$$\mathfrak{m}^3 k[[x,y]] \subseteq I k[[x,y]] + \mathfrak{m}^4 k[[x,y]]$$
and therefore $\mathfrak{m}^3 k[[x,y]] \subseteq I k[[x,y]]$
and so, as $k[x,y]_{(x,y)} \to k[[x,y]]$ is faithfully flat, also $\mathfrak{m}^3 \subseteq I$.
So we have $\mathfrak{m}_o^3 \subseteq (F_o,G_o) \subseteq \local{o}$ and therefore $H_o \in (F_o, G_o)$. As $H(p_i)=0$ we have also $H_{p_i} \in \mathfrak{m}_{p_i}=(F_{p_i}, G_{p_i}) \subseteq \local{p_i}$ for $i=1,\ldots,11$. So locally the preconditions for the AF + BG-theorem are fulfilled for all points $o,p_i$ execpt for possibly $q=p_{12}$.
Now comes the trick: Choose a line $L$ containing $q$ and additionally the points $w_1,w_2,w_3$ on $F$ but not on $G$. Also suppose that $L$ is not a part of $F$, that is $L \nmid F$. Then $L H$ fulfills the preconditions for AF + BG also for the twelfth point $q=p_{12}$.
So we have
$$L H = Q_1 F + Q_2 G$$
Substituting $w_i$ in this equation and noticing $L(w_i) = 0$, $F(w_i) = 0$ and $G(w_i) \neq 0$, we have $Q_2(w_i) = 0$. So the quadric $Q_2$ contains the collinear points $w_1,w_2,w_3 \in L$ and so splits as $Q_2 = L L_1$.
Substituting this, we get
$$L (H - L_1 G) = Q_1 F$$
As $L \nmid F$ was our assumption, we have $Q_1 = L L_2$. So we get
$$L(H - L_1 G) = L L_2 F$$
or, dividing by $L$
$$H = L_2 F + L_1 G$$
which proves $H(q) = 0$.
