1. Convenient Triangle Coordinates

Here I follow closely the book of Andreescu and Andrica, [Andreescu] containing a welth of material handling the
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 <A,A>=a1b1+a2b2 of two
points (or vectors) A(a1,a2), B(b1,b2). The calculations are essentially reduced by selecting the center of coordinates at the
circumcenter O of the triangle of reference ABC. As usual {a,b,c} denote the side-lengths {BC, CA, AB} and A (bold)
denotes the point A(a1,a2) or vector of R2, while A denotes the angle at this vertex of the triangle.

The absolute barycentric coordinates (see BarycentricCoordinates.html ) representing a point Q are three numbers
(xa,xb,xc) with xa+xb+xc = 1 and satisfying the vector equation:
(1)                                                  Q = xaA + xbB + xcC.
The coordinates  (xa,xb,xc) determine the ratios
BA'/A'C = xc/xb,
(*)                       CB'/B'A = xa/xc,
AC'/C'B = xb/xa.
By selecting the origin of the cartesian coordinates at O we have the formulas:
<A,A> = |A|2 = R2,   (and cyclic ...)
<A,B> = R2 -c2/2,        ( cyclic ...)
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 (xa,xb,xc)  (not necessarily satisfying xa+xb+xc = 1) a point Q
through equations (1). The problem with such a "system" is that the association is not one to one and two
tripples (xa,xb,xc) and (xa',xb',xc') may define the same point Q.
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(xa,xb,xc) to
express the point Q they represent through a vector equation.

Lemma
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 (xa',xb',xc') of point Q' = kQ (meant
the scalar multiple of the vector) are given by the equation

To prove this use the two vector equations
Q' = kxaA + kxbB + kxcC.
Q' =  xa'A + xb'B + xc'C.
=>
(xa'-k xa)A+(xb'-k xb)B+(xc'-k xc)C=0.
Last equation implies that
((xa'-k xa), (xb-k xb'), (xc'-k xc)) = q ((B1C2-B2C1), (C1A2-C2A1), (A1B2-A2B1)),
where q is a real number and A=(A1,A2), etc.  By our assumptions on the coordinate system used (B1C2-B2C1)=R2
sin(2A) etc.  The normalization of coordinates implies through summation of the components
1 = k + q R2(sin(2A)+sin(2B)+sin(2C)) ,
the factor of q is seen to be ([Fursenko, p.21])

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

Remark The vector multiplied by (1-k) in the formula above represents the absolute barycentric coordinates of the
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}.

2. G, I, O

Besides the barycentric coordinates of the circumcenter obtained above (see Remark) we can easily compute the
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 xC/XB = c/b etc.. and denoting the half-perimeter by s: 2s = a+b+c,
and normalizing we obtain
I = (a/(2s), b/(2s), c/(2s).

As an example of application of the previous formulas I calculate |IG|2 using the previous expression for the norm.

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|2 = R(R-2r).
Finally computing the norm |G| with the formula of section-1 we find the distance:

3. The GIO triangle construction

Euler solved the problem of constructing a triangle from the points {I,H,O} (H: the orthocenter), which is equivalent to the problem of
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 Fundamental Invariants of the triangle (see Fundamental_Invariants.html ).
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 s2 from the two equations and using again the second of the equations in section-2 to express the radius r:
r = (R2 - OI2)/(2R),                                                                (*)
we obtain (after some easy calculations):

The condition to be satisfied is that  R2 > OI2, which introducing the above expression, Stewart's formula to compute IJ and simplifying leads to:

The last inequality expresses  the aforementioned condition i.e. that the incenter I has to be inside the orthocentroidal circle with diameter GH
and center J. Assuming this is true we obtain the inradius r from Euler's formula   2rR = R2-OI2 and from this the half-perimeter s by solving
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 .

4. The cubic equation satisfied by {a,b,c}

The following argument [Andreescu, p.105] leads to the cubic equation whose roots are the side-lengths of the triangle.
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.

BarycentricCoordinates.html
Fundamental_Invariants.html
EulerRelation.html
OrthicAxis.html
CubicSymmetry.html

Bibliography

[Andreescu] Titu Andreescu and Dorin Andrica Complex Numbers from A to Z Boston, Birkhaeuser 2005
[Fursenko] Fursenko F. B. Lexicographical account of constructional problems of triangle geometry problems Mathematics in school, 1937, no. 5 pp. 4-30, no. 6 pp. 21-45
[Guinand] A. Guinand Euler lines, tritangent centers, and their triangles Amer. Math. Monthly, 91(1984)
[Stern] Joseph Stern Euler's Triangle Determination Problem Forum Geometricorum, Vol. 7(2007)
[YiuEuler] Paul Yiu Conic Solution of Euler's Triangle Determination Problem Journal for Geometry and Graphics Vol. 12(2008), No.1