[alogo] 1. Moebius transformations

Moebius transformations are maps of the complex plane onto itself of the simplest kind. They are defined through a quotient of linear factors which can be expressed by a matrix of real numbers

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The main properties of Moebius transformations are:
[1] They build a group, the composition corresponding to the product of matrices.
[2] They are conformal maps i.e. they preserve angles.
[1] They preserve the cross ratio of four complex numbers.
[3] They map the set of circles and lines to itself.
[5] They are completely determined by three points in general position and its images.

[alogo] 2. Moebius transformations and triangles

Fixing an equilateral triangle A0B0C0 of the plane and using property [5] above, we can define for every other triangle ABC of the plane a unique Moebius transformation F mapping the vertices of the equilateral to corresponding vertices of the triangle ABC. Although this map does not preserves lines it realizes an interesting correspondence between elements of the equilateral and the triangle ABC. This elements have in one or the other way a relation to cross-ratio, which is preserved by this kind of transformations.

Theorem-1
[1] F maps the circumcircle of 0B0C0 to the circumcircle of ABC.
[2] F maps the symmetry axes of A0B0C0 to the Apollonian circles of ABC.
[3] The center of A0B0C0 maps via F to the first isodynamic point J (lying inside the circumcircle) of ABC. The point at infinity maps via F to the second isodynamic point J' of ABC.
[4] The bundle of lines through the center of A0B0C0 maps via F to the bundle of circles (I) passing through the isodynamic points of ABC. The bundle of concentric circles to the circumcircle of A0B0C0 maps via F to the bundle (II) of circles orthogonal to bundle (I) which contains the circumcircle of ABC, the Brocard circle of ABC and its Lemoine axis.


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The proofs are easy consequences of the listed properties of Moebius transformations.
[1]: Follows from the preservation of circles.
[2]: The image of the symmetry axis A0A1 will be a circle k orthogonal to the circumcircle. Taking A1 to be the diametral of A0 we create quadrangle A0B0A1C0 which is harmonic i.e. the cross-ratio (B0, C0, A0, A1) = -1. By the preservation of cross-ratios via F, the intersection point A* of k with the circumcircle of ABC will define also a harmonic quadrangle (B, C, A, A*) = -1. But it is well known that this is the other intersection point of the circumcircle with the symmedian from A, which is a point on the Apollonian circle through A. This identifies k with the Apollonian circle through A.
[3]: Is a consequence of [2], since the isodynamic points are the common points of the Apollonian circles.
[4]: Is a consequence of [3] and the conformality of F. Bundle (II) is the orthogonal one to (I) as this happens by their pre-images through F.

[alogo] 3. A lemma on Moebius transformations

If a Moebius transformation F maps a line/circle (e) to a circle (k), then it conjugates the reflexion/inversion Re with respect to (e) to the inversion Ik with respect to k i.e.

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In the case (e) is a line consider first the point at infinity I and its image J=F(I) by F. Since every line passes through I, its image will pass through J. Thus, circle k=F(e) and the image of a line kz = F(ez), where ez is a line passing through a point z and orthogonal to e, will both pass through point J.
Since F is conformal kz will be a circle or line passing through J, orthogonal to k and passing also through w = F(z). Let now z' be the reflexion of z on e and z0 the middle of zz', which is a point on e. Their images w'=F(z') and w0 = F(z0) are points of the circle/line kz. Besides the cross ratio (I, z0, z, z') = -1 and since F preserves it will be also (J, w0, w, w') = -1. Thus, if kz is a line then {w,w'} are inverse with respect to k. If kz is a circle, then this condition implies that the quadrangle Jww0w' is harmonic, hence, by general properties of harmonic quadrangles the two opposite vertices {w,w'} are inverse with respect to the circle which is orthogonal to the circumcircle of the quadrangle and passes through the other two vertices i.e. {w,w'} are inverse with respect to k as claimed. This proves the case in which (e) is a line.
If (e) is a circle the argument is the same. One has only to consider points {z, z', z0, z1} as before with the one difference, by which z1 is now the other than z0 intersection point of line zz' with the circle (e), replacing the point at infinity I of the previous argument. The figure below shows the relations and the harmonic quadrangle supporting the argument.

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Corollary
In connection with the settings of the previous theorem
[1] The inversions with respect to the Apollonian circles of ABC are conjugate via F to the reflexions on the symmetry axes of A0B0C0.
[2] The inversion w.r. to the circumcircle of ABC is conjugate via F to the inversion w.r. to the circumcircle of A0B0C0.

[alogo] 4. The Brocard and Lemoine axes

In connection with the settings of theorem-1 there is a particular line (e) through the center of the equilateral, which maps via F to the Brocard axis (which is a member of bundle (I)) of the triangle containing the isodynamic points {J, J'} and passing through the circumcenter Q of triangle ABC. There is also a circle (k), concentric to the circumcircle of the equilateral, which maps to the Lemoine axis (which is a member of bundle (II)) of the triangle ABC.

Theorem-2
[1] The pre-images {K1, Q1} via F of the symmedian point K and the circumcenter Q of triangle ABC are two points on (e) symmetric with respect to the center of the equilateral. The circle with diameter K1Q1 maps to the Brocard circle of triangle ABC.
[2] The intersection points {L0, I0} of line (e) and circle (k) map via F respectively to the intersection L of the Lemoine and Brocard axes and the point at infinity I.

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These are consequences of the fact that the Brocard circle, having diameter QK and the Lemoine axis belong to the bundle (II) which is the image via F of the circles centered at O.
Remark Since F respects cross-ratios along (e), L0 is harmonic conjugate to K1 with respect to the intersection points {z,z'} of (e) with the circumcircle of the equilateral (or equivalently, {L0, K1} are inverse with respect to the circumcircle of the equilateral).

[alogo] 5. Brocard points and angle

Theorem-3
[1] The Brocard points {W, W'} of triangle ABC have pre-images under F the other two vertices of the equilateral triangle inscribed in the circle with diameter K1Q1 and having one vertex at Q1.
[2] The Brocard angle of triangle ABC is equal to the angle between line (e) and the circle (f0), which is orthogonal to the circumcircle of the equilateral and passes through points {I0, Q1, W1}.

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The first claim follows from the fact that the circle of bundle (I) passing through {J,J'} and the Brocard point W intersects the Brocard axis by an angle of 120 degrees. Hence its preimage, which is line OW1, makes the same angle with (e). The second claim follows from the fact that the circle (f0) passing through points {I0, Q1, W1} maps via F to line QW.
Remark From the location of the circle (f0) through {I0, Q1, W1} relative to the equilateral Q1W1W2 we see that the Brocard angle is always less than 30 degrees.

[alogo] 6. The Moebius diagram of a triangle

I call Moebius diagram of the triangle the configuration on the left part of the last figure, determined by the following three elements:
(i) The equilateral A0B0C0,
(ii) The line (e) through the center of the equilateral,
(iii) The point I0 on line (e).
Fixing the vertices of the equilateral, it seems that we can parameterize all triangles of the plane by means of pairs (e, I0). In fact, as we saw in the previous sections, every triangle ABC generates such a diagram related to the triangle by the unique Moebius transformation that maps the vertices of the equilateral to corresponding vertices of ABC.
There is though a fine point that should be noticed. Assume that the two triangles A1B1C1 and A2B2C2 define two corresponding Moebius diagrams (e1, I1) and (e2, I2) with respect to the two corresponding Moebius transformations {F1, F2}, mapping the equilateral respectively to these triangles. Obviously the composition

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is a Moebius transformation, which maps one triangle onto the other. G respects also the isodynamic points, since their pre-images under the Fi' s are always the same points, namely the center of the equilateral and the point at infinity. In general though G does not map the Brocard axis of the first triangle to the Brocard axis of the second. In fact, the image of the first Brocard axis is not even a line in the general case.

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The above image illustrates the case. Lines {b1, b2} are the Brocard axes of the respective triangles. Transformation G maps in general b1 onto a circle G(b1) and not onto b2. The (harmonic) quadrangle shown has as vertices the images of the two isodynamic points of the first triangle which are the isodynamic points of the second and the images of the middle L1 and the point at infinity.
There are though cases for which G(b1) and b2 coincide. This happens, for example, when the triangles are similar. This is examined in the file MoebiusDiagram.html .

See Also

MoebiusDiagram.html

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