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I've found a generalisation of Napoleon's theorem to general polygons.

Take any regular $n$-gon inscribed in a circle and stretch it (in any direction) so that the circle becomes an ellipse and the $n$-gon is no longer regular. Then construct regular $n$-gons on the sides of the original $n$-gon. The centroids of these regular $n$-gons make another regular $n$-gon. This is not too hard to prove using vectors and some trigonometric identities. Is this result well-known? If so, is there a nice geometrical reason why it is true?

Hexagons

The case $n=3$ gives Napoleon's theorem because you can get any triangle by stretching an equilateral triangle. The regular $n$-gons in the picture are constructed on the outside. The result applies if they are all constructed on the inside too.

Geogebra link: https://tube.geogebra.org/m/1432065

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  • $\begingroup$ Does it work when you project a circle onto a parabola or a hyperbola? (I think it should, but maybe, you can slightly change your geogebra worksheet a bit to test my question.) $\endgroup$
    – Batominovski
    Jul 23, 2015 at 11:51
  • $\begingroup$ A quick tests suggest that the side length of the final pentagon might be invariant under rotations of the original pentagon inside a circle and depends only on the transformation applied. Have you checked this? $\endgroup$
    – g.kov
    Jul 23, 2015 at 14:34
  • $\begingroup$ Yes, that's definitely true. The final $n$-gon lies on a circle with radius $\cos\left(\frac{\pi}{n}\right) (a+1)$, where the ellipse is $x^2 + a^2 y^2 = a^2$. $\endgroup$
    – octopus
    Jul 23, 2015 at 14:48

2 Answers 2

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This should be the Napoleon-Barlotti theorem.

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Question: If so, is there a nice geometrical reason why it is true?

Solution:

there is a Nice Geometrical reason behind this, so we will explain them with the Help of Three Conjecture.

In given below conjecture we will take Pentagon as an example and those conjecture will also valid if we replace those Pentagon by any $n$ gon.

Conjecture 01: Let $P_1P_2P_3P_4P_5$ be regular Pentagon and Let $Q_1$ be arbitrary point anywhere in Geometry Plane and rotate $Q_1$ about $P_1$ by angle 2π/5 anticlockwise to get $Q_2$ and Rotate $Q_2$ about $P_2$ by 2π/5 anticlockwise to get $Q_3$ and Simillary define {$Q_4$, $Q_5$} cyclically then $Angle$($Q_5P_5Q_1$) =2π/5 as Shown in Figure01

Note for Conjecture 01:

(1) The Pentagon $Q_1Q_2Q_3Q_4Q_5$ lies on Ellipse and have Same Geometrical nature as of Pentagon mentioned above

(2) We named this Pentagon $Q_1Q_2... Q_5$ as Napoleon Barlotti Pentagon so that we could explain next Conjecture in easy manner.

(3) in conjecture 01 ,if we replace angle 2π/5 by 144° then at the end of the construction we will find that $Angle$($Q_5P_5P_1$) =144°.

(4) Simillary the Conjecture 01 wil holds true for any $n$ polygon where we can rotate angle continuously by 2π/n and we suspected that result could be holds true for other angle in place of 2π/n .

Conjecture 02: Let $P'_1P'_2...P'_5$ be translation of $P_1P_2... P_5$ where $P_1P_2... P_5$ is Napoleon Barlotti Pentagon and Center of Regular Pentagon made on Base $P_1P'_2$, $P_2P'_3$ ,...., $P_5P'1$ in Same sence makes Another Regular Pentagon.

Note:

(1) When $P_1$=$P'_1$ then this reduced to Conjecture 01.

(2) the General form of conjecture 02 could be written like this: "Let $P_1P_2... P_n$ be convex irregular $n$gon and $P'_1P'_2... P'_n$ be it's translation and Let $Q_1, Q_2,..., Q_n$ be apex Similar isosceles triangle made on base $P_1P_2$, $P_2P_3$,..., $P_nP_1$ and Let $Q'_1, Q'_2,..., Q'_n$ be apex of Similar isosceles triangle made on Base $P_1P'_2$,$P_2P'_3$..., $P_iP'_i+1$,..., $P_nP'_1$ then any result which will holds true for $Q_1, Q_2,..., Q_n$ will also holds true for $Q'_1, Q'_2,..., Q'_n$ .

In point (2) , For n=5 and isosceles triangle 54°-54°-72° whose apex angle is 72° then this reduced to conjecture 02.

Conjecture 03: Let $P_1P_2P_3... P_10$ be convex irregular 10-gon such that $P_1P_3P_5P_7P_9$ makes regular Pentagon and Simillary point $P_2P_4P_6P_8P_10$ makes another regular Pentagon and Let $Q_1$ be arbitrary point and rotate $Q_1$ about $P_1$ by 2π/5 anticlockwise to get $Q_2$ and rotate $Q_2$ about $P_2$ by 2π/5 anticlockwise to get $Q_3$ and Simillary define {$Q_4$, $Q_5$,..., $Q_10$} cyclically then $Angle$($Q_10P_10Q_1$) =2π/5 as shown in Figure2.0

Note:

(1) The Above Conjecture 03 is Generalisation of Converse of Napoleon Barlotti Theorem so if we go in the converse of Conjecture 03 then we could Generalise the Napoleon Barlotti Pentagon.

(2) Simillary it will holds true for $n$ gon in place of Pentagon.

(3) In case of $n$ gon, we have to do the continuous rotation of 2π/n but we suspected that result could be holds true for other angle also.

[Reference]: We done a lots of work by using Translation, Center of Gravity, Continuous Rotation on Euclid group.io .

Best regards Jayendra jha and Sankalp Savaran.

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  • $\begingroup$ Is our answer is understandable? $\endgroup$
    – Jayendra and Sankalp
    Mar 4 at 2:38

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