271. Relations concerning Fermat points.
by Virgil Nicula, May 3, 2011, 2:38 AM
Relations concerning Fermat points.
Theorem 1. Let
be the apex of the three equilateral triangles constructed on the outside (on the inside) of the triangle 
.
(1) The three lines
are concurrent in a point 
called the first Fermat point (or first isogonic center).
(2) The three lines
are concurrent in a point 
called the second Fermat point (or second isogonic center).
The proof of this theorem is well know (see here).
Theorem 2. Let be the apex of the three equilateral triangles constructed on the outside (on the inside)
of the triangle. If 
, are the first Fermat point, the second Fermat point, the area of the triangle the we have:
[/color]
Theorem 1. Let
be the apex of the three equilateral triangles constructed on the outside (on the inside) of the triangle 
.(1) The three lines
are concurrent in a point 
called the first Fermat point (or first isogonic center).(2) The three lines
are concurrent in a point 
called the second Fermat point (or second isogonic center).The proof of this theorem is well know (see here).
Theorem 2. Let
be the apex of the three equilateral triangles constructed on the outside (on the inside)of the triangle
. If 
, are the first Fermat point, the second Fermat point, the area of the triangle the we have:
[/color]
and 
.
.
we have:
![$R_c = \frac{{c^3 }}{{4\left[ {ABB_1 } \right]}} = \frac{{c^3 }}{{\sqrt 3 c^2 }} = \frac{c}{{\sqrt 3 }}$](http://latex.artofproblemsolving.com/6/a/f/6aff1f02feda4e004e44bdc871655b38d155dfaa.png)
.
and 
,
![$\begin{array}{l}
AF_1 \cdot BF_1 = \frac{{\sqrt 3 \left( {b^2 + c^2 - a^2 } \right) + 4\Delta }}{{2\sqrt 3 m}} \cdot \frac{{\sqrt 3 \left( {b^2 + c^2 - a^2 } \right) + 4\Delta }}{{2\sqrt 3 m}} = \\ \\
= \frac{{\left( {4\Delta + \sqrt 3 c^2 } \right)^2 - \left[ {\sqrt 3 \left( {b^2 - a^2 } \right)} \right]^2 }}{{12m^2 }} = \\ \\
= \frac{{16\Delta ^2 + 3c^4 + 8\sqrt 3 c^2 \Delta - 3b^4 - 3a^4 + 6a^2 b^2 }}{{6\left( {a^2 + c^2 + b^2 + 4\sqrt 3 \Delta } \right)}} \\ \\
\end{array}$](http://latex.artofproblemsolving.com/a/1/2/a126991dba42b5f7731866dfa268498372dfb491.png)
and 
we have
This post has been edited 9 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:24 AM
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