Can you prove or disprove the following claim:
Construct a hexagon circumscribed around a conic section. Intersection points of its non-principal diagonals lie on a new conic section.
GeoGebra applet that demonstrates this claim can be found here.
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$\begingroup$ And the conic section may not necessarily be an ellipse. $\endgroup$– Anindya PrithviCommented Oct 6, 2020 at 8:53
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1$\begingroup$ It is sufficient to prove this when the hexagon is circumscribed about a circle. (This will not necessarily make the derived conic a circle.) The fact that all (non-degenerate) conics are "projectively equivalent" to a circle helps with understanding many such results (for instance, Pascal's Theorem), and exploiting that fact often streamlines proofs. $\endgroup$– BlueCommented Oct 6, 2020 at 9:16
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$\begingroup$ You might consider reposting the heptagon version that you deleted a few months ago. $\endgroup$– brainjamCommented Mar 19, 2021 at 23:39
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1$\begingroup$ @brainjam math.stackexchange.com/q/4069038/15660 $\endgroup$– PedjaCommented Mar 20, 2021 at 5:13
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$\begingroup$ @PeđaTerzić I remember you posted a similar problem about 3 principal diagonals meeting a point and 6 circumcenters lying a conic. Where is the post now? $\endgroup$– auntyellowCommented Sep 22, 2021 at 10:11
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1 Answer
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This is a consequence of Pascal and Brianchon's Theorems.
The intersections of the non-principal diagonals can also be seen as the intersections of the triangles $\triangle{DBF}$ and $\triangle{CAE}$. By Brianchon's theorem, the principal diagonals $EB,FC,AD$ are concurrent at a point $X$. Thus the two triangles are perspective. A converse of Pascal's theorem says that the points of intersection of two perspective triangles lie on a common conic.
Details and more precise statements can be found in Hatton's Projective Geometry, pg 189. There you can find Pascal's theorem, its converse, Brianchon's theorem and proofs.
