# A conic inside a hexagon

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.

• And the conic section may not necessarily be an ellipse. Commented Oct 6, 2020 at 8:53
• 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.
– Blue
Commented Oct 6, 2020 at 9:16
• You might consider reposting the heptagon version that you deleted a few months ago. Commented Mar 19, 2021 at 23:39
• Commented Mar 20, 2021 at 5:13
• @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? Commented Sep 22, 2021 at 10:11

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.