## 22059Re: [EGML] Locus problems

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• Mar 18, 2014

[Antreas]:

Let ABC be a triangle and A'B'C' the cevian triangle of P = O.

The circle (O, OA') intersects again BC at A", the circle

(O,OB') the CA at B" and the circle (O,OC') the AB at C".

1. The parallels through A",B",C" to AA', BB', CC' resp.

concur at H.

2. The triangles ABC and orthic triangle of A"B"C" are perspective (??)

In general, which is the locus of P such that:

1. The parallels through A",B",C" to AA',BB',CC', resp.
are concurrent? (also I is on the locus)

2. The triangles ABC and orthic triangle of A"B"C" are perspective?

[César]:

In general, which is the locus of P such that:

1. The parallels through A",B",C" to AA',BB',CC', resp.
are concurrent at  Q? (also I is on the locus)

M’Cay cubic through excenters and X(1), X(3), X(4),  X(1075), X(1745), X(3362)

For P=I, Q= (b+c)*(a-b+c)*(a+b-c)-a*b*c : : =  (1,3) /\ (4,80)

=  Anticomplement of X(3878) = Reflection of (1/65), (145/3874)

=  ( -0.630855882446927, -0.22633760667943, 4.088524001507182 )

For P=O, Q=H

For P=H, Q=H

2. The triangles ABC and orthic triangle of A"B"C" are perspective at Z?

A circum-sixtic through X(3), X(4)

For P=I, no perspective

For P=O, Z= (94,275) /\ (1144, 4795)  = ( 0.306955496281031, -1.09587885376699, 4.257678074693355 )

For P=H, Z=X(24)

***********************************************

Dear César,

Thanks!

Now, in the same configuration, and for P = I, I will get the points A",B",C"

by another way. That is:

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

A2 = the reflection of A' in BB' (lying on AB)

A21 = the reflction of A2 in AA' (lying on AC)

A213 = the reflection of A21 in CC' (lying on BC)

A3 = the reflection of A' in CC' (lying on AC)

A31 = the reflection of A3 in AA' (lying on AB)

A312 = the reflection of A31 in BB' (lying on BC)

We have A213 = A312 =: A" (= the second intersection of BC and

the circle (I, IA'))

Now, denote:

Hab = the orthocenter of A"A2A21

Hac = the orthocenter of A"A3A31

The line HabHac passes through A" and is parallel to AA'.

Similarly (cyclically) we construct the points

Hbc, Hba and Hca,Hcb.

1. HabHac, HbcHba, HcaHcb concur

(on the OI line, case #1 in the quoted text above)

2. HabHba, HbcHcb, HcaHac concur

(they are parallels, perpendiculars to Euler line)

and by the double perspective theorem we get:

3. HbaHca, HcbHab, HacHbc concur (on the OI line).

APH

:-)

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