barycentric coordinates as weights to combine complex numbers. Here, though, I prefer to avoid complex numbers and

make the calculations using the vector space structure of the plane and use the usual inner product <

points (or vectors) A(a

circumcenter O of the triangle of reference ABC. As usual {a,b,c} denote the side-lengths {BC, CA, AB} and

denotes the point A(a

The absolute barycentric coordinates (see BarycentricCoordinates.html ) representing a point Q are three numbers

(x

(1)

The coordinates (x

BA'/A'C = x

(*) CB'/B'A = x

AC'/C'B = x

By selecting the origin of the cartesian coordinates at O we have the formulas:

<

<

Here R denotes the circumradius, and (cyclic) means that the formulas are valid if we permute the letters {a,b,c}

cyclically. Last equation results from the cosine theorem applied to triangle OAB.

More generally one can associate to each tripple of numbers (x

through equations (1). The problem with such a "system" is that the association is not one to one and two

tripples (x

Anyway, for points represented in this way the norm of the corresponding vector can be expressed by the elements of the

triangle of reference.

The following lemma addresses a subtlety in using the uniquely defined absolute barycentric coordinates(x

express the point Q they represent through a vector equation.

Assume the origin is at the circumcenter O of triangle ABC and write an arbitrary point Q in terms of its absolute

barycentric coordinates as in (1) Then the absolute barycentric coordinates (x

the scalar multiple of the vector) are given by the equation

To prove this use the two vector equations

=>

(x

Last equation implies that

((x

where q is a real number and

sin(2A) etc. The normalization of coordinates implies through summation of the components

1 = k + q R

the factor of q is seen to be ([Fursenko, p.21])

Replacing this in the above equation for q we obtain the claimed formula.

circumcenter O of triangle ABC obtained for the value k=0. The formula itself is then seen to be simply the

expression of the barycentric coordinates of a point on the line XO written as a combination kX + (1-k)O and

consequently as the analogous combination of the barycentric coordinates of {X, O}.

absolute barycentric coordinates of the centroid G and the incenter I using the crucial relations (*).

In fact, the centroid makes these ratios all equal to one giving the coordinates (1,1,1) and normalizing:

G = (1/3 : 1/3 : 1/3).

Analogously the ratios for the incenter are x

and normalizing we obtain

I = (a/(2s), b/(2s), c/(2s).

As an example of application of the previous formulas I calculate |IG|

The cyclic (in a,b,c) sums in this formula depend on the following equations (see Fundamental_Invariants.html ):

Here r denotes the radius of the incircle. Using these we obtain for the varioius sums the following expressions:

Introducing this in the original sum and simplifying (had better done with Maxima to avoid errors) we find:

From Euler's relation (see EulerRelation.html ) between circumradius and inradius we know:

|OI|

Finally computing the norm |G| with the formula of section-1 we find the distance:

constructing the triangle from {G, I, O} (since each triple determines the other). The method can be described as follows.

1) Solve the above three equations with respect to {s,r,R}, called

2) Consider the cubic equation whose roots are the side-lengths {a,b,c} of the trianlge.

3) Solve the cubic ot find these lengths and construct the triangle.

Here I solve the equations with respect to {s,r,R} and in the next paragraph I derive the cubic equation satisfied by {a,b,c}.

Solving the second with respect to 2rR and replacing to the other two equations leads to:

Elliminating s

r = (R

we obtain (after some easy calculations):

The condition to be satisfied is that R

The last inequality expresses the aforementioned condition i.e. that the incenter I has to be inside the

and center J. Assuming this is true we obtain the inradius r from Euler's formula 2rR = R

e.g. the first of the three equations we started with.

The calculations have shown that the location of I inside the orthocentroidal circle is a necessary condition for the solution of the problem. That this

condition is also sufficient has been shown by Guinand [Guinand]. I has though to be different from the middle N of OH which is the center of the

Euler circle ([Stern], [YiuEuler]). For the orthocentroidal circle and related themes see OrthicAxis.html .

The coefficients of the cubic are expressions involving the fundamental invariants {s,r,R} of the triangle.

This cubic equation has {a,b,c} as roots (see CubicSymmetry.html for the symmetry of a cubic about its inflexion

point) and the problem to construct the triangle from its points {G,I,O} is solved by determining the three roots,

which in general leads to a problem not solvable by the classical tools of ruler and compasses.

Fundamental_Invariants.html

EulerRelation.html

OrthicAxis.html

CubicSymmetry.html

[Fursenko] Fursenko F. B.

[Guinand] A. Guinand

[Stern] Joseph Stern

[YiuEuler] Paul Yiu

Produced with EucliDraw© |