Given a family of conics, and a point P, all polars of P with respect to the conic-members of the family pass through another point P' = F(P).

This defines the quadratic transformation (or generalized isogonal transformation or generalized conjugacy) of the family. In practice, P' is constructed as the intersection point of two polars L

A particular case of quadratic transformations are the

[1] Construct the

[2] From the bundle of conics passing through {K,A',B',C'} single out the two degenerate conic-members c

[3] To each point P, associate the intersection point P' of the polars of P with respect to c

Taking the system of points {A,B,C,K} as a

t

Thus K and its harmonic associates {(-1,1,1),(1,-1,1),(1,1,-1)} are the only fixed points of t

The tripolar of K is also easily seen to have the equation:

x+y+z=0.

(1/a)x + (1/b)y + (1/c)z = 0. Thus in every such system the operation of taking the trilinear polar of a point P(a,b,c) has the same expression in coordinates: [(a,b,c) ---> line with coefficients (1/a,1/b,1/c)]. This shows also that the operation is not a projective map (or correlation as is usually named a projectivity between the space of points and the space of lines).

L : ax+by+cz=0

transforms it to a conic passing through the vertices of the triangle.

In fact, setting x' = 1/x, y' = 1/y, z' = 1/z and considering the image of the line (L: ax+by+cz=0) under the transformation, we see that this satisfies

a/x' + b/y' + c/z' = 0 <==> ay'z' + bx'z' + cx'y' = 0,

which is the equation of a conic passing through the vertices of the triangle.

In particular, the trilinear polar of K having equation x+y+z=0 transforms to the conic

(c) : (1/x)+(1/y)+(1/z) = yz+zx+xy=0.

The first form of the equation shows also that a point P(x,y,z) on (c) has a trilinear polar passing through K:

(1/x)+(1/y)+(1/z) = 0 <==> line (1/x)u+(1/y)v+(1/z)w=0 passes through (u,v,w)=(1,1,1).

Another way to say it (see below) is that: conic (1/x)+(1/y)+(1/z)=0 is generated by the tripoles of lines through K.

[1] The (generalized) isogonals t

[2] The trilinear polar tr(P) of a point P(x,y,z) on the conic a/x+b/y+c/z=0 is a line with coefficients (1/x,1/y,1/z), hence passes through point (a,b,c). Equivalently:

[3] The conic consists of the tripoles of all lines through point (a,b,c), which is the isogonal conjugate of (1/a,1/b,1/c) which has ax+by+cz=0 as trilinear polar (see IsotomicConicOfLine.html for a figure illustrating the case).

[1] Let ABC be a triangle and {K,L

[2] Consider also the traces {A

[3] For each line L intersecting the sides of the triangle at points {A',B',C'} respectively lying on {BC,CA,AB} find their harmonic conjugates {A'',B'',C''} with respect to pairs {(A

[4] {A'',B'',C''} are on a line L'. If {A',B',C'} are created by line ax+by+cz=0, then {A'',B'',C''} are created by line (1/a)x+(1/b)y+(1/c)z=0. In other words the tripolar L = tr(P) of the point P(1/a,1/b,1/c) which is ax+by+cz=0, maps by this transform to the tripolar L'=F*(L)=F*(tr(P)) = tr(F(P)) of the isogonal conjugate Q(a,b,c) of P, which is (1/a)x+(1/b)y+(1/c)z=0.

[5] The transformation on lines L'=F*(L) thus defined makes sense for every line not passing through the vertices of ABC. Besides it is involutive and together with the isogonal transformation and the trilinear polar construction L=tr(P) which to P(a,b,c) associates its trilinear polar (1/a)x+(1/b)y+(1/c)z=0 makes the following diagram of transformations commutative.

F* induces a projective map between the two projective lines: P

An element of P

ax

maps via F* to line L': (1/a)x+(1/b)y+(1/c)z=0 which is the trilinear polar of point Q(a,b,c).

Thus, the tripole Q of L'=F*(L) is on line x

In (4) we had also a line L turning around a point P

To simplify matters assume the figure of (1) with coordinate basis {A,B,C,K} and line x+y+z=0 the trilinear polar of K. Then parametrize this line through two points, for example A-B and A-C. Every other point of L

(A-B) -t(A-C) = (1-t)A + (-1)B + (t)C,

which has corresponding trilinear polar

(1/(1-t))x + (-1)y + (1/t)z =0, which is equivalent to (t)x + (-t(1-t))y + (1-t)z=0.

Eliminate t to find the envelope of these lines by taking the derivative:

x + (2t-1)y - z =0 => t = (y+z-x)/(2y).

This, introduced to the previous equation and simplifying reduces it to

-x

It is easy to see that this conic passes through {A

See the file InconicsTangents.html for references to geometric proofs of the same property.

Note that the matrices representing the conics as quadratic forms are respectively:

The matrices have a product which is a multiple of the identity matrix (by the factor 2). Thus the conics they represent are dual to each other (see TriangleConics.html for a related discussion).

All the constructions made in order to define the quadratic transformation or conjugacy are preserved by projective transformations. Thus denoting by {k

P

~

[1] The trilinear polar tr([P]) of [P] is a tangent QP' of the incircle at a point Q of it.

[2] Point Q is the isotomic conjugate of [P] with respect to A'B'C'. Q'=F(Q) is the isotomic conjugate of the same point [P] with respect to ABC.

[3] The trilinear polar tr'(Q) of Q with respect to the medial triangle A'B'C' of ABC passes through the center O and is orthogonal to the direction PB defining the point at infinity [P]. The same is true for the trilinear polar tr(Q') of Q' with respect to ABC. Hence the two trilinear polars coincide. The orthogonal direction to [P] is the direction of the Simson line of Q with respect to A'B'C' or of Q' with respect to ABC.

[4] Lines tr([P]) and tr'(Q) are conjugate in the sense of section 5: t(tr([P])) = tr'(Q) = tr(t'([P])).

[5] Line OP passing through O and the point at infinity [P] has tripole tr(OP) the diametral point of t([P]).

[6] The tangents at {Q,Q'} are parallel, thus determining another point P

The tangency of tr(P) to the incircle follows from section 8. The passing of tr'(Q) (resp. tr(Q')) through O follows from section 3. The identification of the contact point Q with the t'([P]) follows from IncircleTangents.html . From there also we have the parallelity of line tr'(Q) to the Simson line of Q with respect to triangle A'B'C'.

The relations between the elements of ABC and A'B'C' result from the projective invariance discussed in the previous section. The projectivity used is the anti-homothety F sending A'B'C' to ABC. The point at infinity [P] satisfies F([P])=[P], hence t([P])=t(F([P]))=F(t'([P]))=F(Q)=Q'.

The statement on OP is a consequence of the previous statements. Line OP is orthogonal to the Q'-Simson line with respect to ABC and its tripole is the point on the circumcircle of ABC such that the corresponding Simson line is orthogonal to it. According to well known property of Simson lines (see SimsonDiametral.html ) tr(OP) must be the diametral of t([P]).

The statement in [6] follows from the previous statements which relate the directions of the various lines. An easy angle chasing argument shows that t([P]) and t([P

i) The case in which K = I is the incenter. In this case the transformation is the standard

ii) The case in which K = G is the centroid. In this case the transformation is the standard

Note that these transformations are special quadratic

An example in which f

IncircleTangents.html

InconicsTangents.html

IsotomicConicOfLine.html

IsotomicGeneral.html

ProjectiveCoordinates.html

SimsonDiametral.html

TriangleCircumconics.html

TriangleCircumconics2.html

TriangleConics.html

Trilinears.html

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