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I have found this result by exploring for new problems.

If three parabolas share a common directrix and each pair intersect each other in two points, then, the lines joining the two intersection points of each pair of parabolas are concurrent.

The proof is quite simple, so my question is :

Has anyone seen this before? Any reference?enter image description here

Emmanuel José García
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  • Why don't you see about publishing your result in a journal like Forum Geometricorum? The editor will certainly know if the result is new. Link: http://forumgeom.fau.edu/index.html – Alan Jan 23 '14 at 02:30
  • Publishing in that journal is not that easy. I think the editor of FG don't know the result. He just make a comment about the foci and the point of concurrence. I am asking the editor of Jounal of Classical Geometry for references. Thanks. – Emmanuel José García Jan 23 '14 at 03:40

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enter image description here

I found a useful reference and an interesting theorem called the Three Conics Theorem which is similar to yours, but not the same. Here is a link: http://mathworld.wolfram.com/ThreeConicsTheorem.html , and another reference: "The Seven Circles Theorem and other new theorems" by C.J.A. Evelyn, G.B. Money-Coutts, and J.A. Tyrrell (1974).

As you can see , the three conics intersect at I and J and intersect in pairs at (Q1,P1) , (Q2,P2) , (Q3,P3) . The three segments, Q1P1, Q2P2 , Q3P3 are concurrent at X. This picture is taken from the book mentioned above, (there is a similar diagram at mathworld).

Alan
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This is a particular case of Neville: The common chords of three focus-sharing conics meet in the vertices of a quadrangle. In the case of three ellipses, only one of the four vertices is real. In the case of three parabolas, it degenerates further. www.aip.de/People/deliebscher/Publikationen/DreiEllipsen.ps

D.-E.Liebscher
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