1.1  Description of the problem

Let a₁, b₁, c₁, k₁, a₂, b₂, c₂, k₂, a₃, b₃, c₃, k₃, be real numbers such that

a12 + b12 = a22 + b22 = a32 + b32 = 1.

We call d₁, d₂ and d₃ the straight lines of equations

ì ï í ï î k1 a1 x1 + k1 b1 x2 = k1 c1 k2 a2 x1 + k2 b2 x2 = k2 c2 k3 a3 x1 + k3 b3 x2 = k3 c3

Note that in the Java Sketches the scaling numbers k₁, k₂ and k₃ are denoted by a, b and c.

The lines form a triangle of vertices S₁ = d₂Çd₃, S₂ = d₃Çd₁ and S₃ = d₁Çd₂, see Figure .

Figure 1: The triangle

We define

x = æ ç è x1 x2 ö ÷ ø ,   c = æ ç ç ç è k1 c1 k2 c2 k3 c3 ö ÷ ÷ ÷ ø ,   A = æ ç ç ç è k1 a1 k1 b1 k2 a2 k2 b2 k3 a3 k3 b3 ö ÷ ÷ ÷ ø

The matrix equation Ax = C has in general no solution. The least squares solution x₀ of the normal equation

AT A x = AT c

is such that

F(x) = F(x1,x2) = ||A x - c||2 = (xT AT - cT)(Ax - c)

is minimum for x = x₀.

We denote by d(x,L) the distance from the point x to line L. If one computes F(x) one gets:

F(x) = k12 (d(x,d1))2+k22 (d(x,d2))2+k32 (d(x,d3))2

I Case k₁ = k₂ = k₃

The function to be minimized is

F(x) = (d(x,d1))2+(d(x,d2))2+(d(x,d3))2

Let us call l₁ = S₂ S₃, l₂ = S₃ S₁ and l₃ = S₁ S₂, the lengths of the sides of the triangle defined by the lines d₁, d₂ and d₃. The point x₀ such that F(x) is minimum for x = x₀ is the barycenter of S₁, S₂ and S₃ with masses l₁², l₂² and l₃²:

x0 =  l12 l12+l22+l32 S1+  l22 l12+l22+l32 S2+  l32 l12+l22+l32 S3

Problems

1. If the triangle defined by the lines d₁, d₂ and d₃ is isosceles (let us say S₁ S₂ = S₁ S₃), on what line do you expect to find x₀?

2. If the triangle is equilateral where is x₀?

3. The distances of x₀ to the sides of the triangle are proportional to the lengths of the corresponding sides to a certain power r:

d(x0,d1) l1r =  d(x0,d2) l2r =  d(x0,d3) l3r

Guess what is the value of r?

Note: If r = -1, the point x₀ would be the center of gravity; if r = 0, the point x₀ would be the center of the inner circle.

4. If the triangle is rectangular, e.g. d₂ ^d₃, on what line do you find x₀ (see Figure ?

Figure 2: Rectangular triangle

H K2 S1 S3 =  H K3 S2 S1  since the triangles are similar

II Case k₁, k₂ k₃ any real numbers