13488Re: [EMHL] Mittenpunkt And Concurrency Of Three Euler Lines
- Jul 1, 2006Dear Antreas and All My Friends,
Here are some results for the case:
Q=incenter I and fixed, line is OH, P is any point on the line connected Gergonne point and Mittenpunkt point. The concurrent point of three Euler lines as Pe.
(These results may be helpful for general case)
In the triangle PBcCb denote:
circumcircle as (Oa)
orthocenter as Ha,
altitude foot from Bc as B'c
altitude foot from Cb as C'b.
Similarly with another triangles:
PCaAc we have (Ob), Hb, C'a, A'c,
PAbBa we have (Oc), Hc, A'b, B'a.
Denote some intersections of lines:
Ah = AbA'b /\ AcA'c
Bh = BcB'c /\ BaB'a
Ch = CaC'a /\ CbC'b
1. Six points: B'c, C'b, C'a, A'c, A'b, B'a are on one conic (C1)
2. The circle (Oa) is tangent with line AbAc. Similarly for (Ob), (Oc). Other than P, these three circles cut each other at three point: Ap = (Ob) /\ (Oc), similarly define Bp, Cp. Denote circumcircle of ApBpCp as (Op).
3. Six points: Ha, Hb, Hc, Ah, Bh, Ch are on one conic (C2).
4. Pe is one intersection of circle (Op) and conic (C2).
Some special nice cases:
- When P is mittenpunkt then conic (C1) is a circle.
- When P is centroid then conic (C2) is a circle (Op).
Bui Quang Tuan
Quang Tuan Bui <bqtuan1962@...> wrote: Dear Antreas,
I have got a very nice first result:
- Q is fixed and is incenter I
- Line is OH
The locus (or at least all points on this line) is line connected mittenpunkt and Gergonne point.
So mittenpunkt is NO VERY SPECIAL CASE.
Bui Quang Tuan
Quang Tuan Bui wrote:
Very nice idea! I try only some special case and we should do together to get some general results.
I am trying with first case: Q is fixed, and line is OH. I see the locus may be nice: one line?
The mittenpunkt may be is one very special case: Q = incenter I and the locus is only one point: mittenpunkt or some points?
These conjectures may be the good reasons for our efforts.
Bui Quang Tuan
PS: Please note that these only conjectures. I can confirm only with my first mittenpunkt message.
Antreas P. Hatzipolakis wrote:
On 1-07-06, Quang Tuan Bui wrote (partly):
> Given triangle ABC, mittenpunkt Mp, incenter I. One line passingDear Tuan
> through Mp perpendicular to IA cuts lines AB, AC at Ab, Ac respectively.
> Similarly define Bc, Ba, Ca, Cb.
> 4. The most interesting:
> Three Euler lines of triangles MpBcCb, MpCaAc, MpAbBa are concurrent
> at one point P.
We have here two interesting locus families
Let ABC be a triangle, and Q,P two points.
The perpendicular to QA through P intersects AB,AC at Ab,Ac, resp.
Similarly Bc,Ba, and Ca,Cb.
Which is the locus 1. of P 2. of Q
such that the OH lines (Euler Lines) of PBcCb, PCaAc, PAbBa
Case 1 : Q = Fixed Point, P = Variable Point
Case 2 : Q = Variable Point, P = Fixed Point
Special Cases : P, or Q = I,O,H,G,K, etc
And of course we can ask for concurrence of other than OH lines
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