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
$x= \frac {1} {6 {\lambda}}[3(a^2-b^2+c^2) +4 \sqrt {3}{\Delta} ]$
$y= \frac {1} {6 {\lambda}}[3(a^2+b^2-c^2) +4 \sqrt {3}{\Delta} ]$

******************************************************************************************************
Example.
When $a=3, b = 4, c = 5$ , we have
$\Delta = 3 \times 4 / 2 = 6$ and $\lambda = \sqrt {25+12 \sqrt {3}}$ .
$x=\sqrt {25+12 \sqrt {3}}$ ,
$y= \frac {9+4 \sqrt {3}}{\sqrt {25+12 \sqrt {3}}}$ ,
$z=\frac {4\sqrt{3}}{\sqrt {25+12 \sqrt {3}}}$ .
*******************************************************************************************************

This post has been edited 4 times. Last edited by luimichael, Nov 29, 2020, 1:44 AM
Reason: Typo

Algebra and Geometry

Synthetic Division for linear and higher degree divisors

by luimichael, Jan 23, 2020, 3:12 PM

The synthetic division for linear divisor is basically the same as the Horner's Method.
It can be extended to the general case with divisor of higher degrees.

Polynomials

The three distances problem revisited

by luimichael, Mar 25, 2019, 3:34 PM

P is a point inside a square ABCD. The distance from P to the vertices A, B and C are a, b and c respectively.
The problem is to determine the length of a side of the square.
Here is a Geometric approach:
Let $PD =d$ . Then $d^2+b^2=a^2+c^2$ $\implies$ $d^2=a^2+c^2-b^2$ . (by British Flag theorem)
Now, we rotate triangle PCB about the point B through 90 degrees so that BC is moved to BA , while P is moved to P'.
Note that triangle PBP' is a right-angled isosceles triangle with two sides equal to b, and triangle PAP' is a triangle with sides a, b and c.
The area of quadrilateral PAP'B is therefore $\frac { b^2} 2 + Heron(a,\sqrt 2 ,b,c)$
We can apply rotations of figure to triangle PDC, triangle PAD and triangle PBA one by one, and get the followings:
$\frac {c^2} 2 + Heron(b, {\sqrt 2} c, d)$
$\frac {d^2} 2 + Heron(c,{\sqrt 2} d, a)$
$\frac {a^2} 2 + Heron(d,{\sqrt 2}a ,b)$

The sum of the above four expressions actually represent the double of the area of the given square ABCD, and hence the length of a side of ABCD can be found readily.

This post has been edited 1 time. Last edited by luimichael, Sep 14, 2020, 3:42 PM
Reason: Typo

Dual Problems

by luimichael, Mar 3, 2019, 3:43 PM

A better understanding of the three problems:
Problem 1
Fermat point and the minimum of the sum of distances from the three vertices of a given triangle.
Problem 2
Finding the length of an equilateral triangle given the distances from a point to its vertices.
problem 3
A special system of equations
$x^2+xy+y^2 = a^2$
$y^2+yz+z^2=b^2$
$z^2+zx+x^2=c^2$ .

They are basically the same problem in different interpretations! :idea:

This post has been edited 3 times. Last edited by luimichael, Mar 6, 2019, 8:29 PM
Reason: Colour of words changed

IMO Preliminary Selection Contest Hong Kong

by luimichael, Mar 3, 2019, 3:36 PM

https://www.hkage.org.hk/en/competitions/detail/4942

Mathematics Contest

A challenging geometry problem

by luimichael, Apr 11, 2017, 1:01 PM

As shown in the figure, $\angle ABD=20^\circ$ , $\angle DBC=60^\circ$ , $\angle DCA = 50^\circ$ , $\angle ACB=30^\circ$ .
Determine $\angle ADC$
Solution by Trigonometry
Still don't know how to solve this problem by Geometry.

Edit: Problem already solved by sunken rock
See https://artofproblemsolving.com/community/c4h1427621_without_using_trigonometry

Attachments:

This post has been edited 2 times. Last edited by luimichael, Apr 12, 2017, 12:11 AM
Reason: Update

Tower of Nine

by luimichael, Apr 6, 2017, 12:32 PM

Find the last two digits of $9^{9^9}$ and $9^{9^{9^9}}$ .
Solution
Remark
In view of the above proof, we see that the last two digits of the tower of 9 (to any order > 2) will be 89