In Coolidge's venerable *Treatise on algebraic plane curves*, Theorem 28 p. 392 states:


* Let $S=\{P\_1,\ldots ,P\_{10}\} \subset \mathbb{P}^2$ such that for any $i$, there is a sextic curve (**edit**: integral) singular along $S\smallsetminus P\_i$; then there is a sextic curve singular along $S$.


I am probably missing something, but it seems to me that the text before and after this statement has nothing to do with the Theorem. Does someone know a proof?



Rational sextic plane curves
Let me try to reconstruct Coolidge's argument at pages 390-392.


**Note:** there is a misprint at line 4 of p. 392, where "with double points" should be "with triple points".



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> **Proposition 1 [Coolodge, p. 390-391].** Assume that $P\_1, \ldots, P\_8$ are points on the plane, not three on a line and not six on a conic. Let $\phi\_1$, $\phi\_2$ be a basis for the pencil of cubics through the $P\_i$ and let $f$ be a non-degenerate sextic with the $P\_i$ as double points. Then a point $P\_9$ is a further double point of a non-degenerate sextic with double points at $P\_1, \ldots, P\_8$ if and only if it is a smooth point of the curve $$\frac{\partial(\phi\_1, \, \phi\_2, \, f)}{\partial(x\_1, \, x\_2, \, x\_3)}=0.$$
> This curve has triple points at each of the points $P\_1, \ldots, P\_8$ and passes simply through the $12$ singular points of the nodal curves of the pencil of cubics spanned by $\phi\_1$ and $\phi\_2$.
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> **Proposition 2 [Coolidge, Theorem 27, p. 391].** Assume that nine points of the plane are the base locus of a pencil of nodal sextics. Then there are precisely $12$ sextics in the pencil having a further double point.
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Using these results, we can now prove the



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> **Theorem.** Assume that $P\_1, \ldots, P\_{10}$ is a set of ten points such that for any nine of them there is a non-degenerate sextic (and hence a pencil of non-degenerate sextics) having nodes there . Then there is a non-degenerate sextic having nodes at all of the ten points.
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**Proof.** Consider the nodal sextics through $P\_1, \ldots P\_8$. Since by assumption there is a nodal sextic through $P\_1, \ldots, P\_9$, we deduce that $P\_9$ is a smooth point of the curve $C$ of degree $9$ and with triple points at $P\_1, \ldots, P\_8$ described in Proposition $1$. Analogously, consider the nodal sextics through $P\_1, \ldots, P\_7, \, P\_9$. Then $P\_8$ is a smooth point of the curve $D$ of degree $9$ and with triple points at $P\_1, \ldots, P\_7, \, P\_9$.


The $12$ nodal points $Q\_1, \ldots, Q\_{12}$ of the $12$ singular curves of the pencil of nodal sextics through $P\_1, \ldots, P\_9$ (see Proposition 2) must belong to both $C$ and $D$. Moreover, since $$9\cdot 9 - 7 \cdot 3 \cdot 3 - 3 \cdot 1 - 1 \cdot 3 = 12,$$ the only intersection points of $C$ and $D$ are $P\_1, \ldots, P\_9$ and $Q\_1, \ldots, Q\_{12}$.


Finally, by assumption we also have pencils of nodal sextics through $P\_1, \ldots, P\_8, \, P\_{10}$ and through $P\_1, \ldots, P\_7, \, P\_9, \, P\_{10}$. Then $P\_{10}$ belongs to the intersection of $C$ and $D$, too. By the remark above, this means that $P\_{10}=Q\_i$ for some $i$.


Summing up, there is a curve in the pencil of nodal sextics through $P\_1, \ldots, P\_9$ which has a further node at $P\_{10}$. This concludes the proof.


