# "Another Generalisation of Napoleon Theorem" . How to prove this?

EDIT:

[Introduction]:

In geometry, Napoleon's theorem states that if equilateral triangles are constructed on the sides of any triangle, either all outward or all inward, the lines connecting the centres of those equilateral triangles themselves form an equilateral triangle.

[Generalisation of Napoleon Theorem] :

Let A1A2A3A4A5A6 be Convex Hexagon and

Let B1,B2,B3,B4,B5,B6 be apex of equilateral triangle made on base {A1A2, A2A3,A3A4,A4A5,A5A6,A6A1} either Inwardly and outwardly ,

then G(∆B1B2B3); G(∆B3B4B5);G(∆B5B6B1) makes equilateral triangle.

See

Note: When point A1 concide with A2; A3 coincide with A4; A5 coincide with A6 Simultaneously then we get Napoleon Configuration.

[OUR ATTEMPTS]:

we try to prove it by co ordinate geometry and apply the formula to get the apex of equilaterial triangle. Then we tried to add X and Y co ordinate to get Centroid and after that the formula and manipulation become very complex.

But currently we are studying vector Geometry and complex numbers and try to learn how to prove things algebrically.

[Our Question]:

How to prove this Generalisation? Is there is any easy method or elementary geometrical method to prove this?

[Some important Note]:

(1)here Convex Hexagon means : both regular and irregular Hexagon & symbol G(∆ABC) denotes centroid of ∆ABC.

(2)in above Generalisation the description is correct but in image there is a little bit typo mistake in nameing the point {B1,B2,B3,B4,B5,B6} but I hope you will understand it very easily by reading the description.

[REFERENCE]:

(1)We first published our Result on 9 July 2021 on Euclid messenger group

(2)then we published the results on mathsoverflow but that question was delated because we don't have any experience about how to write good question at that time but results was published by rose jolie on this this website (9 month ago)

• Hello friends , here Convex Hexagon means : both regular and irregular Hexagon & symbol G(∆ABC) denotes centroid of ∆ABC. Hope it helps in understanding it.
– user999691
Dec 1, 2021 at 16:56
• My name is jayendra Jha , i along with my friends Sankalp savaran find this. We first mentioned about on Mathsoverflow but as there was no diagram so it was removed and we thanked to Henry for editing it and we also thanks to the person who mentioned our Generalisation here :wlord.org/… . Many thanks to you all.
– user999691
Dec 1, 2021 at 19:07
• Dear Geometers , in above Generalisation the description is correct but in image there is a little bit typo mistake in nameing the point {B1,B2,B3,B4,B5,B6} but I hope you will understand it very easily by reading the description . If you have any doubts and questions then you can ask through email which I have mentioned in my profile.
– user999691
Dec 5, 2021 at 17:09
• Please edit the question to add any context to the body; comments are easily overlooked and may be hidden. ... As for a hint: Consider each vertex as a complex number. Looking at your image, you can write, say, $B_6=A_1+w(A_2-A_1)$, where $w:=\cos60^\circ+i\sin 60^\circ$; this says that $B_6$ is the $60^\circ$ counter-clockwise rotation of $A_2$ about $A_1$. You can express the other $B_k$ likewise. Then the centroids –say, $P$, $Q$, $R$, in counter-clockwise order— are averages of the appropriate vertices. "All you have to do" is show that $R=P+w(Q-P)$. (It may help to know that $1-w+w^2=0$.)
– Blue
Dec 8, 2021 at 10:45
• A link to a published article that gives purely geometric proofs to this result is available at the bottom of this URL: dynamicmathematicslearning.com/new-napoleon-generalization.html Aug 17, 2022 at 6:15