[alogo] 1. The Moebius diagram of a triangle

I call Moebius diagram of the triangle the configuration of three elements consisting from:
(i) An equilateral triangle A0B0C0,
(ii) A line (e) through the center of the equilateral,
(iii) A point I0 on the line (e).

A Moebius diagram results by defining the Moebius transformation F mapping the equilateral A0B0C0 to an arbitrary triangle ABC, in the sense that F maps vertices {A0, B0, C0} to corresponding vertices {A, B, C}.
By such a transformation line (e) is the pre-image of the Brocard axis of ABC and I0 is the pre-image of the point at infinity.
Here I show that such a diagram uniquely characterises the similarity class of a triangle.

For this I fix the equilateral and its circumcircle c0 and consider also all triangles ABC inscribed in the given circle c. Since the study is modulo similarities, this does not imposes a severe restriction on the generality. Most of the preliminary work has been done in the file Moebius.html .

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There it was shown that F maps the inverse Q1 of I0 with respect to c0 to the circumcenter Q of ABC. It was also shown that the symmetric K1 of Q1 with respect to O maps to the symmedian point K of ABC. There was shown also that the symmetry axes of the equilateral triangle map via F to the Apollonian circles. Hence the angle (fi) between such a symmetry axis and the line (e) is equal to the angle between the corresponding Apollonian circle and the Brocard axis. This implies that the triangles corresponding to various angles (fi) but with constant distance of I0 from O correspond to the equi-brocardian triangles inscribed in c.

Theorem-1 Every triangle A'B'C' inscribed in the circle (c) and sharing the same Brocard angle with ABC defines a Moebius diagram (e', I0) resulting by rotating the diagram (e, I0) by a certain angle about O.

The theorem results immediately from the fact that the equi-brocardian triangles inscribed in (c) result by the same Moebius transformation F, by applying it to an other equilateral A1B1C1 inscribed in c0. All these triangles share the same Brocard axis, Brocard points and isodynamic points. Thus their corresponding line e' coincides with e, whereas the corresponding point I0' is at the same distance from O. By turning the whole configuration about O by the angle A1OA0 we identify A1B1C1 with the triangle A0B0C0 and prove the claim.

[alogo] 2. Correspondence to similarity class

Theorem-2 Two triangles are directly (not reversing orientation) similar if and only if they define the same Moebius diagram.

The theorem follows as a corollary of the previous one, since similar triangles have the same Brocard angle and each similarity class is unique (up to rotations) isncribed in (c) among its equi-brocardians. But a direct proof is equally easy. In fact, if triangles ABC and A'B'C' are similar and the corresponding Moebius transformations are F and F', then


is a Moebius transformation between the triangles coinciding with the similarity on the vertices of ABC, which is also a Moebius transformation. Hence it coincides wholly with the similarity and we have


from which follows immediately that the Moebius diagram defined from F and F' coincide.
Inversely, if the Moebius diagrams coincide and {F, F'} are the corresponding Moebius transformations, then the transformation S defined by the previous formula is also a Moebius transformation fixing the point at infinity, since S(I) = F'(F-1(I)) = F(I0) = I. Hence it is a similarity.

[alogo] 3. Determination of the transformation

Theorem-3 A Moebius diagram completely determines the similarity class of the triangle ABC and the Moebius transformation mapping the equilateral to a representative triangle of this class.

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In fact, point I0 determines also the Brocard angle of the triangle. For this construct first Q1 which is the inverse of I0 with respect to c0. Take then the symmetric K1 with respect to O and construct the equilateral triangle Q1W1W2 inscribed in the circle with diameter K1Q1 and with one vertex at Q1. By the analysis made in the aforementioned reference, triangle Q1W1W2 maps via F to the triangle having vertices the two Brocard points and the circumcenter Q of the triangle ABC. The Brocard angle is given by the angle between line (e) and the circle passing through points {I0, Q1, W1}.
To construct the Moebius transformation F, define an isosceles QWW' with angle at the apex twice the Brocard angle calculated above. Then define F as the unique Moebius transformation mapping the vertices of Q1W1W2 to the vertices of QWW' correspondingly. To define the triangle apply F to the vertices of the equilateral A0B0C0.

Remark Given the circle c to which to inscribe the triangle with given diagram the construction of the triangle can be carried out also without the use of the Moebius transform which relates it to the diagram.
In fact, denoting by {z,z'} the diametral intersections of c0 with e, the cross ratio g = (z, z', K1, I0) transfers via F to (w, w', K, I), where I is the point at infinity and {w,w'} the diametral points on the Brocard line. But (w, w', K, I) equals the oriented ratio Kw/Kw', from which the position of K on ww' can be determined. The same remark applies to the isodynamic point J whose position can be also determined by the ratio Jw/Jw', which is equal to the cross-ratio (z, z', O, I0).
Then, using the angle (fi) the vertex A can be located by drawing the circle passing through the isodynamic point J, orthogonal to c and making an angle (fi) with ww' and taking its intersection point with c. This is the Apollonian circle of the triangle ABC passing through A. The other Apollonian circles can be constructed similarly and define through their intersections with c the other vertices of the triangle ABC.

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