• I ve just realised that Bernard s nice observation that OKN is self-polar with respect to the Kiepert Hyperbola is very easily deducible from the properties of
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I've just realised that Bernard's nice observation that OKN
is self-polar with respect to the Kiepert Hyperbola is very easily
deducible from the properties of Px,Qy,Exy that I described recently.

This confirms me in feeling that those facts are in some way fundamental,
so I thought it might be a good idea to start from the beginning, so that
everyone can follow.

Erect isosceles triangles of the same base angle x on each edge
of ABC (the sign convention being that they are outwards if x is a
positive acute angle). Then the triangle A^x B^x C^x formed by their
apices is in perspective with A B C, the perspector being the point P^x
with coordinates
1 1 1
P^x ~ ( ----- : ----- : ----- ),
SA+Sx SB+Sx SC+Sx

where Sx means S.cotx, where S is twice the area of ABC.
This entails that (aa+bb-cc)/2 = SC (&c), and I customarily
abbreviate Sx.Sy...Sz to Sxy...z.

The isogonal conjugate of P^x is

Q^x = ( aa(SA+Sx) : bb(SB+Sx) : cc(SC+Sx) ),

and these coordinates show that as x varies, the locus of Q^x is
the straight line joining

O = ( aaSA : bbSB : ccSC ) and K = ( aa : bb : cc );

that is to say, the Brocard meridian line (or axis), m. The locus
of P^x is therefore the isogonal conjugate of m, which is clearly
the conic through A B C and the isogonals H and G of O and K.
Since it passes through H, this must be a rectangular hyperbola, and
it's called the Kiepert hyperbola.

Now my observations are that if x+y+z is zero (mod pi), then the points

Px, Py, Qz, Exy

are collinear, where Exy is the point H + 3G.cotxcoty on the Euler line.
The proof is an easy calculation. These observations entail many of the
particular values of these points.

I'll discuss a few examples. First, it's obvious that E(-x,-y) = Exy,
and so we get

"The lines Px Py and P(-x) P(-y) meet on the Euler line"
&
" ........ Px Qy ... P(-x) Q(-y) .......ditto......... ".

Next, taking y = x:

"The tangent (to the KH) at x passes through Q(-2x)."

Let's discuss the particular cases of Px and Qx (writing angles in
degrees): clearly we have

P0 = G, P90 = H, P(+-60) = Fn,Fs, P(+-30) = Nn,Ns, P(+-45) = Pn,Ps

Here the Fermat points Fx are the perspectors of the triangles formed by
the apices of (outer and inner) equilateral triangles on ABC, the Napoleonic
points Nx are the perspectors of "Napoleon's triangles" formed by the
centers of these, and the Pythagorean points Px the perspectors of the
centers of squares on ABC rather than triangles.

The corresponding Qx are the isogonals of these, in particular

Q0 = K, Q90 = O, Q(+-60) = In,Is (the isodynamic poles).

We see for instance that the tangents at G and H both pass
through Q0 = K, so that the pole of the Euler line GH is K.
The polar of O is therefore the line joining K to the harmonic
inverse of O in the segment GK, namely the Ninecenter N,
which proves Bernard's assertion that OKN is a self-polar triangle
with respect to the KH.

In fact the tangents at Pn,Ps = P(+-45) meet at Q90 = O,
showing that the line Pn Ps is also the polar of O, whence these
points lie on KN.

The tangents at P(+30) = Nn and P(-60) = Fs meet at Q(+60) = In,
by the way - I don't know if anyone yet pointed that out (the polar of
In is the line Nn Fs).

I was going to add a lot more, including some discussion of conics,
but have to log off soon, so will postpone that.

John Conway
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