Although the simson line is usually introduced using perpendiculars, it can be described
in purely perspective terms. This leads to an alternative approach to pivotal isocubics.
We recast the original definition as follows. The foot of the perpedicular from P to the
side BC of ΔABC is the intersection of AB and the line through P perpendicular to BC.
The latter is the line through P parallel to OM, where O is the circumcentre, and M the
mid-point of BC. M is the vertex on BC of the cevian triangle of the centroid G. Now we
note that lines are parallel if and only if they meet on L∞, the line at infinity i.e. the
tripolar of G or the persectrix of ΔABC and the cevian triangle of G. We now have
a purely projective description.
Definition 1
Suppose that C is a conic and U a point not on C. Let L be the polar of U with
respect to C, and F the tripole of L. Let ΔA'B'C' be the cevian triangle of F.
Let A",B",C" be the intersections of UA',UB',UC' with L. For any P other than U,
let A*,B*,C* be the intersections of PA" and BC, of PB" and AC, of PC" and AB.
ΔA*B*C* is the (C,U)-pedal triangle of P.
By applying a (possibly complex) projective transformation to the figure we can make
C and U into the circumcircle and O. Thus we have our first result.
Theorem 1
If U is a point not on a circumconic C, then the (C,U)-pedal triangle is
degenerate if and only if P lies on C.
Definition 2
When the (C,U)-pedal triangle is degenerate, we call it the (C,U)-simson line of P.
Of course, all projective properties of the simson configuration will be preserved.
In particular, the tripoles of the simpson lines will give a pivotal isocubic K(C,U) with
node
F. Note that F-isoconjugation is a purely projective notion. If P is on
K(C,U),
let Q be the second intersection of PU and C. Then the simson lines of P and Q will
meet on a conic through the vertices of the
cevian triangles of F and of iF(U). Also
the tripoles of the simson lines will be F-isoconjugates, and the line PQ will be a
tangent to the pivotal conic.
We will want barycentric coordinates for later work, so we will give an algebraic proof
of these and other facts.
Theorem 2
Suppose that the circumconc C has perspector P=[p,q,r], and that U =[u,v,w]
does not lie on C. Let F = [f,g,h] be the tripole of L, the C-polar of U.
Then the tripoles of the (C,U)-simson lines of points on C generate the
pivotal isocubic cK(#F,R), where R = [uf/p,vg/q,wh/r].
Proof
As the polar of U is Σ(qw+rv)x = 0, F = [1/(qw+rv)].
The trace of F on BC is A' = [0,g,h], so line UA' has equation
(rv-qw)x-gy+hz = 0. This meets L at A" = [-1,qwg,rvh].
Take Q = [α,β,γ] on C, so p/x+q/y+r/z = 0. Then QA" has equation
(***)x+ (γ+αrvh)y - (β+αqwg) = 0 - the first coefficient is irrelevant.
This cuts BC at A* = [0,gβ(u/α-w/γ)/q,-hγ(v/β-u/α)/r] as Q is on C.
Let us write this as [0,β",γ"], and put α" = fα(w/γ-v/β)/p. Then the points
B* = [-α",0,γ"], C* = [α",-β",0]. These lie on L":x/α"+y/β"+z/γ" = 0.
This demonstrates the collinearity of A*,B*,C* when Q is on C.
The tripole is [α",β",γ"], which we rewrite for the moment as
D* = [fα2(wβ-vγ)/p,gβ2(γu-αw)/q,hγ2(αv-βu)/r].
We know the tripoles will lie on a conico-pivotal isocubic with node F.
The cubic with root [X,Y,Z] has equation ΣXx(yh-zg)2 = 0.
If we substitute the coordinates of D*, we get
(1) KΣXpα2(wβ-vγ)L2/f = 0, where
K = (fgh/pqr)2, and
L = -γβ2ru+αβ2rw+αγ2qv-βγ2qu
= αβ2rw+αγ2qv-u(βγ)(r/γ+q/β)
= (rw/γ2+qv/β2+up/α2)α(βγ)2, as D on C
Thus (1) becomes MΣXp(wβ-vγ)/f, with M symmetric in all variables.
This vanishes for all D on C if and only if the terms in α,β,γ vanish
independently. This leads to the conclusion that
X = uf/p, Y = vg/q, Z = wh/r.
relationships between points P,U,F,R
We defined F in terms of P,U, and derived R. In fact any two determine the others.
We use standard constructs as defined in the Glossary of ETC.
Observe that F is the cevapoint of P and U.
Thus P and U are F-Ceva conjugates, so [p] = [u(-u/f+v/g+w/h)].
Then R = [uf/p] = [f/(-u/f+v/g+w/h)], the U-cross conjugate of F.
So that F is the crosspoint of R and U
In terms of barycentric multiplication, we can see that:
Given U,P : F = P/c(U/P), R = U/c(U/P).
Given F,R : U = Rc(R/F), P = Fc(R/F).
Given P,F : U = Pa(P/F), R = Fa(P/F).
It is easy to verify that R lies on PF.
the generalised simson transform
Definition In the above notation,
For Q = [x,y,z] ≠ F, the generalised simson transform of Q is given by
gs(Q) = [x(w/z-v/y)f/p,y(u/x-w/z)g/q,z(v/y-u/x)h/r].
This generalises the Gibert-Simson transform mentioned in ETC(#2394).
Maple confirms that gs(Q) lies on cK(#F,R) if and only if Q is on C, on a certain
quartic,
or on one of the lines AU,BU,CU. The quartic Q has equation
rwx2y2+puy2z2+qvz2x2-2pqrxyz(x/f+y/g+z/h) = 0.
It degenerates into a sideline and a cubic if any of the points [-f,g,h],[f,-g,h],[f,g,-h]
lie on C.
Also, the quartic Q is easily seen to be the F-isoconjugate of a conic.
Definition
If U does not lie on the conic C, then for Q on C, the (C,U)-antipode of Q
is the second intersection of C with QU. It is Q itself when QU is a tangent.
Theorem 3
In the above notation, if Q is on C and Q* is the (C,U)-antipode of Q,
then gs(Q) and gs(Q*) are F-isoconjugate.
The (C,U)-simson lines of Q,Q* meet on the conic C(F,R) of Proposition 1
Proof
Suppose that Q = [α,β,γ] is on C, so p/α+q/β+r/γ = 0.
The line QU is Xx+Yy+Zz = 0, where X = βw-γv, Y = γu-αw, Z = αv-βu.
Then it is easy to check that Q* = [p/αX,q/βY,r/γZ] lies on QU, and on C
as Q is on QU. We rewrite Q* as [p/(w/γ-v/β),q/(u/α-w/γ),r/(v/β-u/α)].
The first coordinate of gs(Q*) is
(f/(w/γ-v/β))(w(v/β-u/α)/r-v(u/α-w/γ)/q)
= -(f/α(w/γ-v/β))(vwp/qr+uw/q+uv/r)), as Q is on C
= -pf/α(w/γ-v/β)S,
where S = vwp+uwq+uvr is symmetric and non-zero as U is not on C.
This is the first coordinate of the F-isoconjugate of gs(Q).
For the last part, now that we know that gs(Q) and gs(Q*) are F-isoconjugates,
we observe that the calculation of the intersection of their tripolars is identical
to the perspector calculation of Proposition 1.
If U is outside C, we have two points Q on C where QU is tangent to C.
Then we easily check that, as Q = Q*, [α] = [p/(w/γ-v/β)], so that
gs(Q) = F. These points Q are the intersections of C with the polar of U,
i.e. the tripolar of F and with the quadric Q above.