22020RE: 2 points made from the concurrency of GK lines

Expand Messages
• Dec 1, 2013
• 0 Attachment

Dear friends,

Sorry, I forgot adding the denominator of the above barycentrics.

The denominator for x =3(b^2+c^2-a^2)+2S'

Best regards,

Seiichi.

---In Hyacinthos@yahoogroups.com, <anopolis72@...> wrote:

---------- Forwarded message ----------
From: Seiichi Kirikami
Date: Sat, Nov 23, 2013 at 5:47 PM
Subject: [EGML] 2 points made from the concurrency of GK lines
To: Anopolis@yahoogroups.com

Given a triangle ABC and the point X(13), we denote by Ba and Bc the intersections of the parallel line through X(13) with AB and CA respectively. The other intersections Cb, Ab, Ac and Bc are defined cyclically. GK lines of X(13)AcAb, X(13)BaBc and X(13)CbCa concur in a point Q, where G and K mean the centroid and the symmedian point of ABC respectively.

If we use X(14) instead of X(13), we have another concurrent point R.

Barycentrics for X(13) is as follows.

x=8a^16+(-23(b^2+c^2)+4S’)a^14-(16(b^4+c^4)-73b^2c^2+26(b^2+c^2)S’)a^12+(134(b^6+c^6)-84b^2c^2(b^2+c^2)+46(b^4+c^4+b^2c^2)S’)a^10-(185(b^8+c^8)-34b^2c^2(b^4+c^4)-123b^4c^4+2(b^6+c^6+30b^2c^2(b^2+c^2))S’)a^8+(77(b^10+c^10)+52b^2c^2(b^6+c^6)-91b^4c^4(b^2+c^2)-(72(b^8+c^8)-86b^2c^2(b^4+c^4)+6b^4c^4)S’)a^6+(38(b^12+c^12)-132b^2c^2(b^8+c^8)+159b^4c^4(b^4+c^4)-130b^6c^6+(74(b^10+c^10)-74b^2c^2(b^6+c^6)+6b^4c^4(b^2+c^2))S’)a^4-(b^2-c^2)^2(44(b^10+c^10)-27b^2c^2(b^6+c^6)-17b^4c^4(b^2+c^2)+(26(b^8+c^8)-4b^2c^2(b^4+c^4)+4b^4c^4)S’)a^2+(b^2-c^2)^4(11(b^8+c^8)+9b^2c^2(b^4+c^4)-40b^4c^4+(2(b^6+c^6)-20b^2c^2(b^2+c^2))S’).

S’ means 2*sqrt(3)* area of the triangle.

kx=0.0368160852136.. (non-ETC).

Barycentrics for X(14): replace S’ in baricentrics for x(13)by –S’.

kx=3.15628332574…(non-ETC).

Best regards,

Seiichi Kirikami

• Show all 4 messages in this topic