composition h = g*f is a homothety with center E lying on the line AB and ratio t given by the product of the ratios t = r*s.

The position of the composite homothety center E on line AB is determined by the ratio k = EA/EB = (s-1)/[s(1-r)]

(both homotheties assumed to be different from the identity i.e. non-trivial).

Apply Menelaus to triangle ABC with secant the line XD joining an arbitrary point X and its composite image g(f(X))=D, with C=f(X):

(EA/EB)*(DB/DC)*(XC/XA) = 1.

DB/DC = -BD/DC = -BD/(BC-BD) = -1/((BC/BD)-1) = -1/((1/s)-1) = s/(s-1).

XC/XA = -XC/AX = -(AC-AX)/AX = 1-AC/AX = 1-r =>

EA/EB = (s-1)/[s(1-r)].

This shows the independence of the position of point E from the particular X and its images f(X), g(f(X)).

The ratio ED/EX can be deduced by applying again Menelaus theorem to the triangle EAX with secant the line BC:

(BE/BA)*(CA/CX)*(DX/DE) = 1.

BE/BA = EB/AB = EB/(EB-EA) = 1/(1-EA/EB) = 1/(1-(s-1)/[s(1-r)]) = s(1-r)/(1-sr).

CA/CX = AC/XC = AC/(AC-AX) = r/(r-1) =>

DX/DE = [(1-sr)/s(1-r)] [(r-1)/r] => ED/EX = rs. This shows the claim about the ratio t of the composite homothety.

behind (if the radii are not equal and one does not count the inverse homothety). One has positive ratio, the other has negative.

The first homothety f

directed radii. The ratio is r

and inversely directed radii. The ratio can be set r

harmonic conjugate with respect to {A,B}.

homothety centers.

selecting two of them in all three possible ways.

1) The six homothety centers created by the recipe of the previous section are aligned by three. This follows immediately from the previous

section, since a composition of any two homotheties (out of the six) defines a homothety of the same kind.

2) The homothety centers are divided in two sets {B

collinear points. The line on which they lie is the trilinear polar L (see TrilinearPolar.html ) of a point D. Point D is the common point on lines

{A

3) The trilinear pole (tripole) of line L with respect to triangle B

ratio, is a point E which describes a circle with center D as B

Note that triangles A

at infinity. Consequently, by Desargues theorem (see Desargues.html ) they are also line-perspective and this perspectivity-line is precisely L.

The translation is along the vector AB, which is determined by the centers of the two homotheties.

This can be easily seen by inspecting the relations in the triangle with vertices {A, B, X'=F(X)} created by a point and its image under the first

homothety F

AB and in ratio X'X''/AB = (r-1)/r. Thus X'' = X + [(r-1)/r]AB.

is true if we interchange the order and build a composition of a Translation and a Homothety. Thus, taking the union of homotheties and translations

we have a group.

The figure shows both compositions T*F and F*T of a homothety F

and X' = F(X) is its image under the homothety, X'' = T(X') is the translation of X' by v. Point Y = T(X) and Y' = F(Y). The fact is that the resulting

compositions are homotheties with centers on the line through A, which is parallel to v, but they are different, hence the two transformations are not

commutative. The order by which we apply them plays a role.

The composition T*F (first F, then T) maps X to X'', such that XX'/XA = 1-r =XX''/XB = X'X''/AB. Thus AB = [1/(1-r)]v determines the location of

B and the ratio BX''/BX = AX'/AX = r is the same with that of the homothety F.

The other composition F*T maps X to Y', such that AY'/YY' = r/(1-r) = XY/CA. Thus CA = [(1-r)/r]v determines again the center of the homothety.

The ratio CY'/CX is again r i.e. the same with that of the initial homothety.

at a particular point P

T*F with the same ratio at P

Selecting in particular as P

where r is a non-zero number and {u,v} arbitrary numbers. The matrix represents such a composition of a homothety (at the origin) and a translation

and its inverse is.

relating to section-3. Let ABC a triangle and draw three parallels {k,m,p} to its sides intersecting them at points { K,L,M,N,P,Q} as shown in the

figure below. The the intersection points {K*, M*, P*} of pairs of lines {(BC,MQ), (CA,PK), (AB,NL)}, if they exist, are collinear.

The three triangles created {AKL, MBN, QPC} are pairwise homothetic and the points in question are among their homothety centers. For example

MBN and QPC are homothetic with respect to K*. The proof reduces to that of section-3 by using the circumcircles of these triangles.

TrilinearPolar.html

Desargues.html

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