- --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:

> Let ABC be a triangle, Oa,Ob,Oc and Ha,Hb,Hc the

I think that the lines O'aHa, O'bHb, O'cHc

> circumcenters, orthocenters, resp. Of IBC,ICA,IAB, resp.

> and O'a,O'b,O'c the reflections of Oa,Ob,Oc

> in BC,CA,AB, resp.

>

> Are the lines O'aHa, O'bHb, O'cHc concurrent?

are parallels to AI,BI,CI, resp.

through the vertices A',B',C' of the

cevian triangle of the isotomic conjugate of I.

Anyway, we have a locus problem:

Let ABC be a triangle, P a point, P*

its isotomic conjugate, and A1B1C1, A2B2C2

their cevian triangles.

Let La,Lb,Lc be the parallels to AA1, BB1, CC1

through A2,B2,C2, resp.

Which is the locus of P such that La,Lb,Lc are

concurrent? The whole plane??

APH - Dear Antreas and other Hyacinthists,

this configuration is very interesting, but huge amounts of results

are really popping out when I slightly modify your idea. See below.

First, your conjectures:

> Let ABC be a triangle, Oa,Ob,Oc and Ha,Hb,Hc the

Yes, they are. As you already pointed out, O'aHa and OaI are parallel.

> circumcenters, orthocenters, resp. Of IBC,ICA,IAB, resp.

> and O'a,O'b,O'c the reflections of Oa,Ob,Oc

> in BC,CA,AB, resp.

>

> Are the lines O'aHa, O'bHb, O'cHc concurrent?

To see why this is so, one could for example observe that IHa =

2R.cos(BIC), where R is the circumradius of BIC.

On the other hand, we can easiley see that Oa is the midpoint of arc

BC in the circumcircle of ABC. It is well known that OaB = OaC = OaI =

R, so we can now quickly see that indeed OaO'a = 2R.cos(BIC) = IHa.

Because OaO'a and IHa are parallel, the conclusion follows.

The point of concurrency in the above problem is the Nagel point of ABC.

> Let ABC be a triangle, P a point, P*

Indeed! The lines are concurrent at the anticomplement of P, which is

> its isotomic conjugate, and A1B1C1, A2B2C2

> their cevian triangles.

>

> Let La,Lb,Lc be the parallels to AA1, BB1, CC1

> through A2,B2,C2, resp.

>

> Which is the locus of P such that La,Lb,Lc are

> concurrent? The whole plane??

obvious if one considers the anticomplementary triangle A'B'C'.

La,Lb,Lc clearly pass through A',B',C' and the conclusion follows.

This applies in the above case, the Nagel point is the anticomplement

of I.

After checking all your cases, only the following are whole plane

concurrent:

> 1. O'aHa, O'bHb, O'cHc

Concurrent at the Nagel point

> 2. O'aGa, O'bGb, O'cGc

Concurrent! I have no idea at which point though.

> 3. O'aNa, O'bNb, O'cNc

Concurrent at I.

It's easy to see that O'a is the circumcenter of BCHa. This triangle

however has Na as it's nine point center, so O'aNa is the Euler line

of BCHa, which passes through it's orthocenter I.

> 10. N'aOa, N'bOb, N'cOc

Yes! Very nice, these three lines are concurrent at the inverse of I

in the circumcircle.

_______________________________________________________

So 4 of your 12 triples were concurrent, that's a nice score!

But I think I can do better ;-)

Why, instead of considering the reflections of these centers in AB,BC

and AC, not considering the reflections of these centers in the

MIDPOINTS of AB,BC,AC.

So for example H'a,H'b,H'c are the reflections of Ha,Hb,Hc in the

midpoints of BC,AC,AB.

I think if we do this, ALL the mentioned lines are concurrent.

Some special interesting cases:

4) H'aOa, H'bOb, H'cOc

H'a,H'b,H'c are the excenters of ABC in this case, so the point of

concurrence is I.

10) N'aOa, N'bOb, N'cOc

The point of concurrency is the anticomplement of the Feuerbach point

_______________________________________________________

Further questions:

- In general, what can be said about the 12 points of concurrency we

met just now?

- I strongly conjecture the following, is it true?

Let ABC be a triangle and P a point on the circumcircle or Neuberg

cubic of ABC. Call Oa,Ob,Oc and Ha,Hb,Hc the circumcenters and

orthocenters of BPC,APC,APB.

Define Sa(k1),Sb(k1),Sc(k1) and Ta(k2),Tb(k2),Tc(k2) as the points on

OaHa, ObHb, OcHc which divide (OiHi) (i=a,b,c) in the ratio k1 and k2

respectively.

Furthermore, call S'a(k1),S'b(k1),S'c(k1) and T'a(k2),T'b(k2),T'c(k2)

their reflections in the midpoints of BC,AC,AB.

S'a(k1)Ta(k2), S'b(k1)Tb(k2), S'c(k1)Tb(k2) are concurrent.

Sa(k1)T'a(k2), Sb(k1)T'b(k2), Sc(k1)T'b(k2) are concurrent.

- What can be said about the point of concurrency in the above theorem

(if true)?

Best regards,

Jan