## 22014[EGML] 2 concurrent Euler lines made from a triangle and Lemoine circle

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• Nov 8, 2013
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[Seiichi Kirikami]:

Dear friends,

Given a triangle ABC and 1st Lemoine circle(L) with its center of symmedian point(K), we denote the intersection of AB and L nearest to A by Ab and the intersection of AC and L nearest to A by Ac. Define Bc, Ba and Ca, Cb cyclically. Euler lines AAbAc, BBcBa and CCaCb concur in a point P1. Euler lines KAbAc, KBcBa and KCaCb concur in a point P2.

Barycentrics of Pi: x=a^2((a^2+b^2+c^2)(SAB+SBC-2SCA)(SBC+SCA-2SAB)+ib^2(SCA-SBC)(SBC+SCA-2SAB)+ic^2(SAB-SBC)(SAB+SBC-2SCA)).

(Non ETC, i=1, Search x=8.769078041215.., i=2, Search x=4.880993268140..)

Note: SAB=SA*SB.

Best Regards,

Seiichi.

[Peter Moses]:

Seiichi's point with i = 1:
a^2*(c^2*(SA*SB-SB*SC)*(SA*SB-2*SA*SC+SB*SC)+b^2*(SA*SC-SB*SC)*(-2*SA*SB+SA*SC+SB*SC)+(a^2+b^2+c^2)*(SA*SB-2*SA*SC+SB*SC)*(-2*SA*SB+SA*SC+SB*SC))::
= X[6] + 2 X[74]
Search = 8.76907804121589020082598111013181409429
on lines {{3, 67}, {6, 74}, {25, 125}, {64, 1177}, {110, 3796}, {146, 3589}, {165, 2836}, {186, 1503}, {246, 1976}, {343, 3448}, {399, 5092}, {524, 2071}, {895, 3532}, {1204, 1205}, {1597, 2777}, {2453, 2790}, {2916, 2931}, {3516, 5095}}

Seiichi's point with i =2
a^2*(2*c^2*(SA*SB - SB*SC)*(SA*SB - 2*SA*SC + SB*SC) + 2*b^2*(SA*SC - SB*SC)*(-2*SA*SB + SA*SC + SB*SC) + (a^2 + b^2 + c^2)*(SA*SB - 2*SA*SC + SB*SC)*(-2*SA*SB + SA*SC + SB*SC))::
= 2 X[6] + X[74]
Search = 4.88099326814078001307598313897317561065
on lines {{2, 98}, {3, 895}, {4, 1177}, {6, 74}, {54, 67}, {69, 5504}, {113, 3618}, {185, 575}, {186, 2393}, {217, 5038}, {265, 1176}, {389, 1205}, {403, 1503}, {511, 2071}, {576, 1204}, {578, 5095}, {631, 5181}, {1316, 2790}, {2854, 5085}, {2892, 3541}, {3431, 5505}}

Seiichi's point with general i
= i X[6] + (3 - i) X[74]

Peter Moses