## PRIZE (Re: ORTHOCENTER - REFLECTIONS - CONCURRENT CIRCLES)

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• [APH]: In fact we can take any point P (instead of H) and any points O1,O2,O3 on the circumcircles of PBC,PCA,PAB, resp. Then the circumcircles of the
Message 1 of 2 , Sep 14 3:04 AM
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[APH]:

In fact we can take any point P (instead of H) and any points
O1,O2,O3 on the circumcircles of PBC,PCA,PAB, resp.

Then the circumcircles of the triangles

AO2O3, BO3O1, CO1O2

are concurrent.

Anopolis #850
http://groups.yahoo.com/neo/groups/Anopolis/conversations/messages/850

For a proof I offer the book:

R. G. SANGER: SYNTHETIC PROJECTIVE GEOMETRY (1939)

APH

• The problem is equivalent to this: Let ABC, A B C be two triangles. If the circumcircles of A BC, B CA, C AB are concurrent, then the circumcircles of AB C ,
Message 2 of 2 , Sep 14 10:13 AM
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The problem is equivalent to this:

Let ABC, A'B'C' be two triangles.

If the circumcircles of A'BC, B'CA, C'AB are concurrent, then
the circumcircles of AB'C', BC'A', CA'B' are also concurrent .

It sounds old! And for a reference the offer holds as well !!.

Note: From the other version ie
"If P is a fixed point and O1,O2,O3 three variable points on
the circumcircles of PBC,PCA,PAB, resp  then the circumcircles
of AO2O3, BO3O1, CO1O2 are concurrent" we may ask:
As O1,O2,O3 move on the respective  circles, in what plane region
the points of concurrence are located on?
(I think it is not the entire plane)

APH

On Sat, Sep 14, 2013 at 1:04 PM, Antreas Hatzipolakis wrote:

[Attachment(s) from Antreas Hatzipolakis included below]

[APH]:

In fact we can take any point P (instead of H) and any points
O1,O2,O3 on the circumcircles of PBC,PCA,PAB, resp.

Then the circumcircles of the triangles

AO2O3, BO3O1, CO1O2

are concurrent.

Anopolis #850
http://groups.yahoo.com/neo/groups/Anopolis/conversations/messages/850

For a proof I offer the book:

R. G. SANGER: SYNTHETIC PROJECTIVE GEOMETRY (1939)

APH

--
http://anopolis72000.blogspot.com/
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