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## FW: X(5390) = EULER-MORLEY-ZHAO POINT

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• Dear friends, Shi Yong of Shanghai has developed barycentrics for X(5390) and has kindly given permission to share the work. It starts with the Jan 03 message
Message 1 of 1 , Jan 5, 2013
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Dear friends,

Shi Yong of Shanghai has developed barycentrics for X(5390) and has kindly given permission to share the work. It starts with the Jan 03 message copied at the bottom, where the two pictures are identical. So, just use the left one, which constructs X(5390) in the complex plane with (0,0) at F. The Jan 05 message derives barycentrics from the picture. Can someone verify? It would be nice to have a Mathematica program that starts with the barycentrics and shows the result graphically.

Best regards,
Clark Kimberling

From: YooYooXp YooYooXp [mailto:yooyooxp@...]
Sent: Saturday, January 05, 2013 6:08 AM
To: Kimberling, Clark
Subject: Re: X(5390) = EULER-MORLEY-ZHAO POINT

Dear Professor Clark:

If the interior angles of the triangle ABC are u,v,Pi-u-v.

a=exp((Pi-2*u/3)i)=-cos(2*u/3)+i*sin(2*u/3);
b=exp((2*v/3-Pi)i)=-cos(2*v/3)-i*sin(2*v/3);

Through the appropriate zoom, revolve & translation to the triangle ABC,
we can give the following facts.

A=2/(a+1),B=2/(b+1),C=2*(a+b-1)/(a^2+a*b+b^2);
A_=2/(1/a+1),B_=2/(1/b+1),C_=2*(1/a+1/b-1)/(1/a^2+1/a*1/b+1/b^2);

Then agrees w=exp((Pi/3)i)=(1+sqrt(3)*i)/2;

P1=a^3*b^3+a^2*b^4+a*b^4+a^4-a^3*b-a*b^3+b^4+a^2*b+b^3+a*b;
Q1=(a^3*b^2+a^2*b^3+a^3*b+a^2*b^2+a*b^3+a^2+b^2)*(a-b);
P2=a*b^2+a^2-3*a*b+b^2+b;
Q2=(a*b+1)*(a-b);

P1_=1/a^3*1/b^3+1/a^2*1/b^4+1/a*1/b^4+1/a^4-1/a^3*1/b-1/a*1/b^3+1/b^4+1/a^2*1/b+1/b^3+1/a*1/b;
Q1_=(1/a^3*1/b^2+1/a^2*1/b^3+1/a^3*1/b+1/a^2*1/b^2+1/a*1/b^3+1/a^2+1/b^2)*(1/a-1/b);
P2_=1/a*1/b^2+1/a^2-3*1/a*1/b+1/b^2+1/b;
Q2_=(1/a*1/b+1)*(1/a-1/b);

By complex analysis, Zhao & Zhao_ express as follows.

Zhao=2*(P1*w+Q1)/(P2*w+Q2)/(a+1)/(b+1)/(a^2+a*b+b^2);
Zhao_=2*(P1_*1/w+Q1_)/(P2_*1/w+Q2_)/(1/a+1)/(1/b+1)/(1/a^2+1/a*1/b+1/b^2);

For barycentric coordinates, 3 equations are as listed below.

A*K1+B*K2+C*K3=Zhao;
A_*K1+B_*K2+C_*K3=Zhao_;
K1+K2+K3=1;

Solving the system of equations, we can get K1,K2,K3.

K1=(Zhao*b^3-2*b^2+2*b-Zhao_)*(a+1)/2/(b^2-b+1)/(b-a);
K2=(Zhao*a^3-2*a^2+2*a-Zhao_)*(b+1)/2/(a^2-a+1)/(a-b);
K3=(a^2+a*b+b^2)*(2-Zhao-Zhao_)/2/(a^2-a+1)/(b^2-b+1);

So, EULER-MORLEY-ZHAO POINT = X(5390) = (K1:K2:K3).

Respectfully yours,

Shi Yong

2013/1/5 Kimberling, Clark <ck6@...<mailto:ck6@...>>
Dear Shi Yong,

Thank you for sending the constructions. You wrote that if necessary we can also convert from complex involution analysis to barycentric coordinates. Would you be willing to carry out this conversion? There seems to be considerable interest in the Zhao center. Perhaps they could be put into a Mathematica program using various substitutions.

Best regards,
Clark

From: YooYooXp YooYooXp [mailto:yooyooxp@...<mailto:yooyooxp@...>]
Sent: Thursday, January 03, 2013 2:33 PM
To: Kimberling, Clark
Subject: X(5390) = EULER-MORLEY-ZHAO POINT

[内嵌图片 1][内嵌图片 1]

【HYPERLINK_1】http://fanyi.youdao.com/WebpageTranslate?keyfrom=webfanyi.top&url=http%3A%2F%2Ftieba.baidu.com%2Fp%2F890790231&type=ZH_CN2EN

【HYPERLINK_2】http://tieba.baidu.com/p/890790231

【HYPERLINK_3】http://faculty.evansville.edu/ck6/integer/unsolved.html

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