Re: More problems on the ORTHIAL TRIANGLES configuration

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• Dear Antreas and other Hyacinthists, ... C AcBc. ... X s ... a ,b ,c ... I ll try to give an amazing answer to your question. Let Oa,Ob,Oc the cevian triangle
Message 1 of 2 , Nov 5, 2001
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Dear Antreas and other Hyacinthists,

> [APH]:
> >Let ABC be a triangle. The perpendiculars to AC, AB at A,
> >intersect the BC at Ca, Ba, resp. (ie ACa _|_ AC, ABa _|_ AB).
> >
> >The perpendicular to ACa at Ca, and the perpendicular to ABa at Ba
> >intersect at A'.
> >
> >Similarly define the points B', C'.
> >
> > A
> > /\ ACa _|_ AC, ABa _|_ AB
> > / \
> > / \ A'Ca _|_ ACA, A'Ba _|_ ABa
> > / \
> > / \
> > Ca--------B----------C----------Ba
> >
> >
> >
> >
> >
> >
> >
> > A'
> >
>
> In an old thread, I have called the triangles ABaCa, BCbAb, CAcBc as
> "orthial triangles" of ABC. Let me call now the triangles A'BaCa,
> B'CbAb, C'AcBc as "synorthial triangles" of ABC.
> ....
> Let's now move to synorthial triangles:
>
> Let H'a,H'b,H'c; I'a,I'b,I'c; G'a,G'b,G'c; K'a,K'b,K'c be the
> H's, I's, G's, K's of the synorthial triangles A'BaCa, B'CbAb,
C'AcBc.
>
> Let a'(X), b'(X), c'(X) be the cevians A'X'a, B'X'b, C'X'c of the
X's
> of the synorthial triangles, where X'i = H'i,I'i,G'i,K'i, i =
a',b',c'
>
> Are the following triads of lines concurrent?
>
> 1- a'(H), b'(H), c'(H)
>
> 2- a'(I), b'(I), c'(I)
>
> 3- a'(G), b'(G), c'(G)
>
> 4- a'(K), b'(K), c'(K)

Let Oa,Ob,Oc the cevian triangle of O.
ha = homothecy with center Oa, ratio = -tan(B)tan(C),...
Then A'BaCa = ha(ABC)...
Now, if we consider a point P, the three lines ha(AP),hb(BP),hc(CP)
will concur iff P lies on the line at infinity or on J = Jerabek
hyperbola.
Thus the triangle centers X for which a'(X),b'(X),c'(X) concur are
exactly those who lye on the union of the line at infinity and
Jerabek hyperbola - only H,K in your question -
More over, if P lies on J, the common point of ha(AP), hb(BP), hc(CP)
is the second common point of the line KP and the rectangular
hyperbola through O,K,L and the infinite points of J.
Friendly. Jean-Pierre
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