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
> 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?
Yes, they are. As you already pointed out, O'aHa and OaI are parallel.
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*
> 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??
Indeed! The lines are concurrent at the anticomplement of P, which is
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
After checking all your cases, only the following are whole plane
> 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
- 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
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