- Dear colleagues!
Arsenij Akopjan noted next fact.
Let given three rays OX, OY, OZ and three circles inscribed in angles XOY,
YOZ, ZOX. Then three common interior tangents to this circles distinct from
OX, OY, OZ concur.
We haven't a proof.
- Dear Alexey,
In Hyacinthos message #9463, you wrote:
>> Arsenij Akopjan noted next fact.This is obviously equivalent to the MANNHEIM THEOREM (proof
>> Let given three rays OX, OY, OZ and three circles inscribed
>> in angles XOY, YOZ, ZOX. Then three common interior
>> tangents to this circles distinct from OX, OY, OZ concur.
in Hyacinthos message #6199):
If three circles are given such that the three pairwise
internal common tangents (one of every pair) concur,
then the three other internal common tangents also concur.
- Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.
Ab := PB' /\ BC, Ac := PC' /\ BC
Bc := PC' /\ CA, Ba := PA' /\ CA
Ca := PA' /\ AB, Cb := PB' /\ AB
Oa, Ob, Oc := The Circumcenters of PAbAc, PBcBa, PCaCb
1. For which point(s) AbAc = BcBa = CaCb ?
(I think that for P = (tanA ::) in trilinears)
2. For which point(s) the circumcircles of PAbAc, PBcBa, PCaCb
are congruent? (for P = H)
3. Which is the locus of P such that ABC, OaObOc are perspective?
4. Which is the locus of P such that A'B'C', OaObOc are perspective?