A review on a system of equation

by luimichael, Sep 14, 2020, 4:44 PM

The following system of equation is simple and elegant, and the solution of the system of eqautions actually is closely related the Fermat Point Problem.
I would first solve the system using a methond invoving Algebra and Geometry.

Given that $a,b , c > 0$, solve

$x^2 +y^2 +xy = a^2$---(1)
$y^2+z^2 +yz = b^2$---(2)
$z^2 +x^2 +zx = c^2$---(3)
, where x,y and are positve real numbers.

Solution

We first construct a traingle ABC with side BC=a, CA = b and AB = c.
In the interior of triangle ABC, there is a point F such that $\angle AFB = \angle BFC = \angle CFA = 120^0$.
Let $FB=x , FC = y $ and $FC=z$.
Then triangle ABC is divied into four triangles FAB, FBC and FCA.
Let us denote the area of triagle ABC by $\Delta$.
Consider areas of the four triangles:
$\frac 12 xy \sin 120^0 + \frac 12 yz \sin 120^0 + \frac 12 yz \sin 120^0 = \Delta$
$xy+yz+zx=\frac { 4\Delta }{\sqrt 3}$---(4)
(1)+(2)+(3)+(4)*3:
$2x^2+2y^2+2z^2+4(xy+yz+zx)=a^2+b^2+c^2+{4\sqrt 3} {\Delta}$
$2(x+y+z)^2=a^2+b^2+c^2+{4\sqrt 3} {\Delta}$
$x+y+z=\sqrt{\frac{a^2+b^2+c^2+{4\sqrt 3} {\Delta}}2}$(denote this expression by $\lambda$)
$x+y+z= \lambda$---(5)
(1)-(2): $x^2-z^2 + xy-yz = a^2-b^2$
$(x-z)(x-z)+y(x-z) = a^2-b^2$
$(x+y+z)(x-z) = a^2-b^2$
$\lambda (x-z) = a^2-b^2$
$x-z= \frac {a^2-b^2}{ \lambda}$---(6)
Similarly we have $y -z = \frac {a^2-b^2}{\lambda}$---(7)
(6)+(7): $x+y-2z = \frac {2a^2-b^2-c^2}{\lambda}$
$(\lambda - z)-2z = \frac {2a^2-b^2-c^2}{\lambda}$ by (5)
$3z = \lambda - \frac {2a^2-b^2-c^2}{\lambda}$
$3z = \frac {{\lambda }^2-(2a^2-b^2-c^2)}{\lambda }$
$3z=\frac {\frac{a^2+b^2+c^2+4\sqrt{3}{\Delta}}{2}-(2a^2-b^2-c^2)}{\lambda}$
$z= \frac {1} {6 {\lambda}}[3(-a^2+b^2+c^2) +4 \sqrt {3}{\Delta} ] $
Similarly, we can show that
$x= \frac {1} {6 {\lambda}}[3(a^2-b^2+c^2) +4 \sqrt {3}{\Delta} ] $
$y= \frac {1} {6 {\lambda}}[3(a^2+b^2-c^2) +4 \sqrt {3}{\Delta} ] $

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Example.
When $a=3, b = 4, c = 5$, we have
$\Delta = 3 \times 4 / 2 = 6$ and $\lambda = \sqrt {25+12 \sqrt {3}}$.
$x=\sqrt {25+12 \sqrt {3}}$,
$y= \frac {9+4 \sqrt {3}}{\sqrt {25+12 \sqrt {3}}}$,
$z=\frac {4\sqrt{3}}{\sqrt {25+12 \sqrt {3}}}$.
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This post has been edited 4 times. Last edited by luimichael, Nov 29, 2020, 1:44 AM
Reason: Typo
Algebra and Geometry

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