## Three circles

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• 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
Message 1 of 3 , Mar 4, 2004
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, ... This is obviously equivalent to the MANNHEIM THEOREM (proof in Hyacinthos message #6199): If three circles are given such that the three
Message 2 of 3 , Mar 4, 2004
Dear Alexey,

In Hyacinthos message #9463, you wrote:

>> 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.

This is obviously equivalent to the MANNHEIM THEOREM (proof
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 :=
Message 3 of 3 , Aug 31, 2006
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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