1 NORMALIZING THE LEAST-SQUARES SOLUTION OF LINES

1.1 Description of the problem

Let a_{1}, b_{1}, c_{1}, k_{1}, a_{2}, b_{2}, c_{2}, k_{2}, a_{3}, b_{3}, c_{3}, k_{3}, be real numbers such that

a_{1}^{2} + b_{1}^{2} = a_{2}^{2} + b_{2}^{2} = a_{3}^{2} + b_{3}^{2} = 1.

We call d_{1}, d_{2} and d_{3} the straight lines of equations

\{ \begin{matrix}k_{1}a_{1}x_{1} & + & k_{1}b_{1}x_{2} & = & k_{1}c_{1} \\ k_{2}a_{2}x_{1} & + & k_{2}b_{2}x_{2} & = & k_{2}c_{2} \\ k_{3}a_{3}x_{1} & + & k_{3}b_{3}x_{2} & = & k_{3}c_{3} \end{matrix}

Note that in the Java Sketches the scaling numbers k_{1}, k_{2} and k_{2} are denoted by a, b and c. The lines form a triangle of vertices S_{1} = d_{2}∩d_{3}, S_{2} = d_{3}∩d_{1} and S_{3} = d_{1}∩d_{2}, see Figure 1. We define

Figure 1: The triangle

x =(\begin{matrix}x_{1} \\ x_{2} \end{matrix}), c =(\begin{matrix}k_{1}c_{1} \\ k_{2}c_{2} \\ k_{3}c_{3} \end{matrix}), A =(\begin{matrix}k_{1}a_{1} & k_{1}b_{1} \\ k_{2}a_{2} & k_{2}b_{2} \\ k_{3}a_{3} & k_{3}b_{3} \end{matrix})

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

A^{T}Ax = A^{T} c

is such that

F (x) = F (x_{1}, x_{2}) = ∥Ax − c∥^{2} = (x^{T} A^{T} − c^{T})(Ax − c)

is minimum for x = x_{0}. We denote by d(x, L) the distance from the point x to line L. If one computes F (x) one gets:

F (x) = k_{1}^{2} (d(x, d_{1}))^{2} + k_{2}^{2} (d(x, d_{2}))^{2} + k_{3}^{2} (d(x, d_{3}))^{2}

I Case k_{1} = k_{2} = k_{3}

The function to be minimized is

F (x) = (d(x, d_{1}))^{2} + (d(x, d_{2}))^{2} + (d(x, d_{3}))^{2}

Let us call l_{1} = S_{2}S_{3}, l_{2} = S_{3}S_{1} and l_{3} = S_{1}S_{2}, the lengths of the sides of the triangle defined by the lines d_{1}, d_{2} and d_{3}. The point x_{0} such that F (x) is minimum for x = x_{0} is the barycenter of S_{1}, S_{2} and S_{3} with masses l_{1}^{2}, l_{2}^{2} and l_{3}^{2}:

x̂ = \frac{l_{1}^{2}}{l_{1}^{2} + l_{2}^{2} + l_{3}^{2}}S_{1} + \frac{l_{2}^{2}}{l_{1}^{2} + l_{2}^{2} + l_{3}^{2}}S_{2} + \frac{l_{3}^{2}}{l_{1}^{2} + l_{2}^{2} + l_{3}^{2}}S_{3}

Problems

1. If the triangle defined by the lines d_{1}, d_{2} and d_{3} is isosceles (let us say S_{1}S_{2} = S_{1}S_{3}), on what line do you expect to find x_{0}?

2. If the triangle is equilateral where is x_{0}?

3. The distances of x_{0} to the sides of the triangle are proportional to the lengths of the corresponding sides to a certain power r:

\frac{d(x̂, d_{1})}{l_{1}^{r}} = \frac{d(x̂, d_{2})}{l_{2}^{r}} = \frac{d(x̂, d_{3})}{l_{3}^{r}}

Guess what is the value of r? Note: If r = −1, the point x_{0} would be the center of gravity; if r = 0, the point x_{0} would be the center of the inner circle.

4. If the triangle is rectangular, e.g. d_{2} ⊥ d_{3}, on what line do you find x_{0} (see Figure 2?

Figure 2: Rectangular triangle

\frac{HK_{2}}{S_{1}S_{3}} = \frac{HK_{3}}{S_{2}S_{1}} since the triangles are similar

II Case k_{1}, k_{2} k_{3} any real numbers

For this general case we already made experiments!