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Can you provide a proof for the following claim:

Claim. Construct a convex heptagon circumscribed about an ellipse. Intersection points of its non-adjacent longer diagonals lie on a common ellipse.

enter image description here

GeoGebra applet that demonstrates this claim can be found here.

I have noticed that principal diagonals of the hexagon defined by the points: $𝐺,FG \cap ED,𝐷,𝐶,𝐵,𝐴$ concur at the point $P$. Similar is true for the points $Q,R,S,M,N,O$ as well . So I guess we should apply Brianchon's theorem somehow.

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  • $\begingroup$ It looks like this claim is true not only for a heptagon circumscribed about an ellipse, but also for a heptagon circumscribed about any conic. I only know for sure that, if the conic is a circle, the seven points $M, N, O, P, Q, R$, and $S$ lie on an ellipse. $\endgroup$
    – YNK
    Mar 22, 2021 at 7:25
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    $\begingroup$ Probably Brianchon's theorem and its dual Pascal's theorem may help you to prove this claim. But, you better have a look at Heptagon theorem too. $\endgroup$
    – YNK
    Mar 22, 2021 at 15:05
  • $\begingroup$ I tried the same with an octagon circumscribed to an ellipse: the inner octagon formed by the diagonals is circumscribed to another ellipse. i.stack.imgur.com/Qs1Tt.png $\endgroup$ Mar 22, 2021 at 16:08
  • $\begingroup$ @YNK Thank you for your useful comments. $\endgroup$
    – Peđa
    Mar 22, 2021 at 16:14
  • $\begingroup$ @Intelligentipauca Very interesting. $\endgroup$
    – Peđa
    Mar 22, 2021 at 16:14

1 Answer 1

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The comments mention the Heptagon Theorem:

Let $H$ be a heptagon with seven vertices given in cyclic order inscribed in a conic. Then the Pascal lines of the seven hexagons obtained by omitting each vertex of $H$ in turn and keeping the remaining vertices in the same cyclic order are the sides of a heptagon $I$ which circumscribes a conic.

Moreover, the Brianchon points of the seven hexagons obtained by omitting the sides of $I$ one at a time and keeping the remaining sides in the natural cyclic order are the vertices of the original heptagon.

Since this is a theorem about points, lines, conics, and incidence, it is a theorem of projective geometry and can be dualized. The dual of the first paragraph is

Let $H$ be a heptagon with seven sides given in cyclic order which circumscribes a conic. Then the Brianchon points of the seven hexagons obtained by omitting each side of $H$ in turn and keeping the remaining sides in the same cyclic order are the vertices of a heptagon $I$ inscribed in a conic.

As noted in the OP, the Brianchon points of the seven hexagons are the pairwise intersections of consecutive long diagonals (by which we mean their endpoints are in cyclic order, such as $AD,BE$). So the dualized first paragraph proves the OP's claim.

The Heptagon theorem is stated and proven in Evelyn et al, The Seven Circles Theorem and Other New Theorems (if you can't get access to the book, the proof is shown below). As done above, the Heptagon theorem can be dualized, or the proof of the Heptagon Theorem can be dualized, which is a lot more work.

It's worth pointing out, if it isn't already obvious, that the Heptagon Theorem contains dual constructions. Starting with an inscribed heptagon one uses Pascal lines to construct a circumscribed heptagon. One can then reverse the construction using Brianchon points to construct an inscribed heptagon.

The picture given in the Mathworld article uses a doubly winding blue heptagon to obtain a smaller red heptagon. But, if you start with a singly winding heptagon, you end up with a red heptagon typically much larger than the blue heptagon. For example the Pascal lines of the orange heptagon $MNOPQRS$ in the OP are the sidelines of the blue heptagon $ABCDEFG$.

Below is a copy of the proof of the Hexagon Theorem.

enter image description here enter image description here enter image description here

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  • $\begingroup$ I appreciate it very much if you could add the proof of the heptagon theorem to your answer at your convenience. $\endgroup$
    – YNK
    Mar 24, 2021 at 9:48
  • $\begingroup$ @YNK, I've added the proof. $\endgroup$
    – brainjam
    Mar 24, 2021 at 21:30
  • $\begingroup$ Thank you very much. $\endgroup$
    – YNK
    Mar 25, 2021 at 7:07
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    $\begingroup$ Very interesting. By the way, here is a recent question about a (regular) heptagon... My answer is very awkward but the other answer is nice ! $\endgroup$
    – Jean Marie
    Apr 9 at 21:24
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    $\begingroup$ @JeanMarie, yes, there have been similar questions lately about regular n-gons and the best answers are quite clever. You’d think there would be a general method for these questions but I haven’t seen anything like that. $\endgroup$
    – brainjam
    Apr 9 at 23:14

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