## 1. Product of homotheties

Assume that fA,r, gB,s are two homotheties with centers {A,B} and ratios {r,s} respectively, such that the product  rs is not 1. Then the
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).

Remark This property,  is one of the most interesting applications of Menelaus theorem (see Menelaus.html ).

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.

## 2. Pairs of circles and homotheties

The most natural generation of homotheties is by pairs of circles. Whenever you see two circles {A(r), B(r')} there are two homotheties
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 f1 with positive ratio has center C at the intersection of AB and a line joining the end-points of two parallel and equal
directed radii. The ratio is r1 = r/r'. Second homothety f2 has center D at the intersection of AB with a line joining the end-points of two parallel
and inversely directed radii. The ratio can be set r2 = -r/r'. Both homotheties map circle  B(r') to A(r). The two homothety centers {C,D} are
harmonic conjugate with respect to {A,B}.
Remark In case there are common tangents, symmetrically lying on AB,  their intersection point coincides with one of these two
homothety centers.

## 3. Tripples of circles

Tripples of circles {A1(r1), A2(r2), A3(r3)} combine in interesting relations with the homotheties defined as in the previous section, by
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 {B12, B13, B23} and {C12, C13, C23}. The second tripple consists of
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
{A1C23, A2C13, A3C12 }.
3) The trilinear pole (tripole) of line L with respect to triangle B1B2B3, whose vertices are the orbit of B1 under the set of three homotheties with positive
ratio, is a point E which describes a circle with center D as B1 moves on circle A1(r1).

Note that triangles A1A2A3 and B1B2B3 are point-perspective, since {A1B1, A2B2, A3B3} are parallel by definition, thus meeting
at infinity. Consequently, by Desargues theorem (see Desargues.html ) they are also line-perspective and this perspectivity-line is precisely L.

## 4. The singular case

This is the excluded in section-1 case, in which rs=1. In this case the composition of the two homotheties {F, G} is a translation.
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 FA,r. Then it is easily seen that the image of X' under the second homothety  X'' = G(X')  is on line X'B and such that X'X'' is parallel to
AB and in ratio  X'X''/AB = (r-1)/r. Thus X'' = X + [(r-1)/r]AB.

## 5. Homothety and Translation composition

Could not leave the subject without looking at the composition of a Homothety and a Translation. The result is seen to be a Homothety. The same
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 FA,r and a translation Tv by a vector v. These are applied to a point X
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.

Remark The previous result implies that an arbitrary homothety can be represented as a composition of a translation and a homothety centered
at a particular point P0. In fact, if F is a homothety of ratio r at another point P, then the translation T with vector v = (1-r)PP0 defines a homothety
T*F with the same ratio at P0.
Selecting in particular as P0 the origin of coordinates we can represent the group of Homotheties + Translations as a group of matrices:

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.

## 6. Application

To pay a tribute to Yaglom, whose book is the best reference for the material discussed here, I include the following exercise [Yaglom, II, p. 33]
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.