# Does there exist a curve of degree 11 having 15 triple points?

Does there exist an irreducible curve of degree 11 in the projective plane which would have 15 triple points?

For information, such a curve would be rational, if it exists, and would be smooth at all other points (compute the genus with the classical formula: $(d-1)(d-2)/2-\sum a_i(a_i-1)/2=45-3\cdot 15=0$)

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if we start from three lines with a common point, then add two lines through some point of that triad, we have degree 5 and only two triple points, so adding 6 more lines gives only 14 triple points. thus even if we can smooth the extraneous double points we lose. maybe one should pass some conics through more nodes. we could pass one conic through all 4 nodes at step 3 above, getting a degree 7 curve with 6 triple points, and 4 nodes. passing another conic through all 4 gives a degree 9 curve with 10 triple points, and 10 nodes. so we get a reducible one, then try to smooth it. / – roy smith May 31 '12 at 2:24

According to the paper, the virtual dimension of the linear system is $78-90-1<0$ since the degree of the equation is at least $5$. So it looks like the answer is no. – Will Sawin May 31 '12 at 3:00