Hyacinthos : Message: Conjecture

http://anthrakitis.blogspot.com/2012/02/pp-pp.html
<http://anthrakitis.blogspot.com/2012/02/pp-pp.html>

(it is too late to check if I have already posted it before!)

APH

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Wed Feb 22, 2012 11:28 pm

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"Antreas" <anopolis72@...>

xpolakis

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Dear Hyacinthians Could someone check my conjecture? Let ABC be an acute-angled triangle.Then r(4R + r) >= sqrt(3)S , S is the area Merry Christmas, ...	martin lukarevski martinton	Dec 21, 2001 9:25 pm
Dear Martin, [ML]: Could someone check my conjecture? Let ABC be an acute-angled triangle.Then r(4R + r) >= sqrt(3)S , S is the area *** This is item...	Paul Yiu yiuatfauedu	Dec 21, 2001 9:43 pm
Dear Paul, [ML]: Could someone check my conjecture? Let ABC be an acute-angled triangle.Then r(4R + r) >= sqrt(3)S , S is the area [PY]: This is item...	martinton	Dec 22, 2001 7:59 pm
http://anthrakitis.blogspot.com/2012/02/pp-pp.html <http://anthrakitis.blogspot.com/2012/02/pp-pp.html> (it is too late to check if I have already posted it...	Antreas xpolakis	Feb 22, 2012 11:28 pm
Dear Antreas and Chandan, according to "Let Q be a point on the line PP* and P1P2P3 the circumcevian triangle of Q with respect the triangle A1B1C1 ...", the...	Francisco Javier garciacapitan	Feb 23, 2012 9:21 pm
Dear Francisco, ... Your circle is the pedal circle of P and P* and obviously Q=P1=P2=P3. Best regards Nikos Dergiades...	Nikolaos Dergiades ndergiades	Feb 24, 2012 8:07 am
It is true. Here is a proof:- Suppose, A3B3C3 is the circumcevian triangle of P* wrt A2B2C2. R be the reflection of Q on O. A4B4C4 be the circumcevian triangle...	Chandan Banerjee cbanerjee.2011	Feb 23, 2012 8:36 pm
Let R1R2R3 be the reflection of P1P2P3 in PP* The triangles A2B2C2, R1R2R3 are perspective. http://anthrakitis.blogspot.com/2012/02/pp-pp.html ...	Antreas xpolakis	Feb 24, 2012 10:41 am
Suppose, PP* intersects the circumcircle of A1B1C1 at U,V. A2U and A2V are the bisectors of \angle P1A2R1. So (A2P1,A2R1;A2U,A2V)=-1. So A2R1 passes through...	Chandan Banerjee cbanerjee.2011	Feb 25, 2012 8:55 pm

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