## An equilateral triangle in plane geometry

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• Dear Geometers, Let ABC be a triangle in plane geometry, A0B0C0 be the excentral triangle of ABC, A1B1C1 be the excentral triangle of A0B0C0, Define: AiBiCi be
Message 1 of 3 , May 31, 2017

Dear Geometers,

Let ABC be a triangle in plane geometry, A0B0C0 be the excentral triangle of ABC, A1B1C1 be the excentral triangle of A0B0C0, Define: AiBiCi be the excentral triangle of A{i-1}B{i-1}C{i-1} with i=1,2,3,...Then show that AnBnCn be an equilateral triangle when n \to  \infty

Best regards

Sincerely

Dao Thanh Oai

• Dear Geometers, I give a proof of the conjecture: An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+2^(n+1)]/2^(n+1)C/3
Message 2 of 3 , May 31, 2017
Dear Geometers,

I give a proof of the conjecture:

An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+2^(n+1)]/2^(n+1)C/3

Bn=[(-1)^(n-1)+2^n]/2^nB/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3+2^(n+1)]/2^(n+1)A/3

Cn=[(-1)^(n-1)+2^n]/2^nC/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3+2^(n+1)]/2^(n+1)B/3

So when n--> \infity then An=Bn=Cn=180^0/3=60^0

The perspector of AnBnC with ABC?

Best regards
Sincerely
Dao Thanh Oai
• Dear Geometers, I am sorry, I eidted a typo I give a proof of the conjecture:
Message 3 of 3 , May 31, 2017
Dear Geometers,

I am sorry, I eidted a typo

I give a proof of the conjecture:

An=[(-1)^(n-1)+2^n]/2^nA/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3

Bn=[(-1)^(n-1)+2^n]/2^nB/3+[(-1)^n+2^(n+1)]/2^(n+1)C/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3

Cn=[(-1)^(n-1)+2^n]/2^nC/3+[(-1)^n+2^(n+1)]/2^(n+1)A/3+[(-1)^n+2^(n+1)]/2^(n+1)B/3

So when n--> \infity then An=Bn=Cn=180^0/3=60^0

The perspector of AnBnCn with ABC?

Best regards
Sincerely
Dao Thanh Oai
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