- 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.

Sincerely Alexey - 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.

Sincerely,

Darij Grinberg - 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?

Antreas

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