Numerical solution to Fermat Point Problem

by luimichael, Nov 3, 2007, 9:38 AM

Given a traingle ABC with BC = a, CA = b and AB = c.
Determine a point P on the plane of ABC such that PA + PB + PC is minimum.
Solution:
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Assume the case that each of the angles of ABC is less than 120 degrees.
Then we have the result that the point P should be located inside ABC such that
angle APB = angle APC = angle BPC = 120 degrees.
Under such assumption, let PA = x, PB = y and PC = z.
By Cosine Formula, $x^2 + y^2 - 2xy \cos 120^o = c^2$ , which simplifies to
$x^2 + y^2 + xy = c^2$ -

(1)
Similarly we have
$y^2 + z^2 + yz = a^2$ -

(2)
$z^2 + x^2 + zx = b^2$ -

(3)
Denote $\Delta$ as the area of triangle ABC.{readily found by Heron's formula}
Sum of areas of triangles PAB , PBC and PCA = Area of ABC:
$\frac 1 2 xy \sin 120^o + \frac 1 2 yz \sin 120^o + \frac 1 2 zx \sin 120^o = \Delta$ .
$xy + yz + zx = \frac { 4 \Delta }{\sqrt 3}$ -

(4)
(1)+(2)+(3)+3*(4):
$2x^2 + 2y^2 + 2z^2 + 4xy + 4yz + 4zx = a^2 + b^2 + c^2 + 4 \sqrt 3 \Delta$
$2(x + y + z)^2 = a^2 + b^2 + c^2 + 4 \sqrt 3 \Delta$
$x + y + z = \sqrt {\frac {a^2 + b^2 + c^2 + 4 \sqrt 3 \Delta} 2}$ -

(5)

Next Question:
Find the values of PA, PB, PC separately for which their sum is minimum.

Solution:
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Denote $\lambda = \sqrt {\frac {a^2 + b^2 + c^2 + 4 \sqrt 3 \Delta} 2} = x + y + z$
(1)-(2): $x^2 - z^2 + x - z = c^2 - a^2$
$(x - z)(x + y + z) = c^2 - a^2$
$x - z = \frac {c^2 - a^2}{\lambda}$ -

(6)
Similarly, (1) -(3) implies
$y - z = \frac {c^2 - b^2}{\lambda}$ -

(7)
(5) -(6)-(7):
$3z = \lambda - (\frac {2c^2 - a^2 - b^2}{\lambda})$
$= \frac {{\lambda}^2 - 2c^2 + a^2 + b^2}{\lambda}$
$= \frac {2{\lambda}^2 - 4c^2 + 2a^2 + 2b^2}{2\lambda}$
$= \frac {4\sqrt 3 \Delta + 3(a^2 + b^2 - c^2)}{2\lambda}$
Therefore,
$z = \frac {4\sqrt 3 \Delta + 3( a^2 + b^2 - c^2)}{6\lambda}$
$x = \frac {4\sqrt 3 \Delta + 3( - a^2 + b^2 + c^2)}{6\lambda}$
$y = \frac {4\sqrt 3 \Delta + 3( a^2 - b^2 + c^2)}{6\lambda}$

Example:
For a = 3, b = 4 and c = 5.
$\Delta = 3*4/2 = 6$ since the triangle is right-angled.
$\lambda = \sqrt {25 + 12\sqrt 3}$
By substitution,
$x \approx 3.388521647$ ,
$y \approx 2.354003099$ , and
$z \approx 1.023907822$
These values satisfy the equations (1), (2) and (3) by direct checking.

Further question:
Despite the fact that the three equations (1), (2) and (3) are related to concrete Geometric entities, does the solution above represent the algebraic solution of the system (1),(2) and (3)?
In other words, is that the solution by radical ?
Can it be applied in the case when a, b and c are Complex numbers ?

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Numerical solution to Fermat Point Problem

by luimichael, Nov 3, 2007, 9:38 AM

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