Τετάρτη, 18 Ιανουαρίου 2012

PAUL PRIZE


Paul Erdos was a great mathematician who used to pose problems in mathematics periodicals (American Mathematical Monthly and others) with money prizes. I will do the same for a problem, but instead of money, the prize will be a classical book of geometry. It is a good coincidence that I have named the Point of the problem by the name of another great Paul: Pablo Picasso, and also today (18 January) is the birthday of my son Paul.

The Problem:

Let ABC be a triangle, A'B'C' the antipodal triangle of ABC (=circumcevian triangle of O) and D a point. Let A"B"C" be the circumcevian triangle of D with respect A'B'C' (ie A" is the other than A' intersection of A'D with the circumcircle etc)


Denote:

N,N1,N2,N3 the NPC centers of the triangles ABC, A"BC, B"CA, C"AB.

The four centers N,N1,N2,N3 are concyclic.

The center of the circle is General Picasso Point [See HERE]

Terms of Proofs:

The proof must be Euclidean synthetic (ie with no use of algebra or coordinate geometry)

The Prize:

The first, who will post a proof to the list Hyacinthos, will win the book:

F.G.-M.: Exercices de Geometrie. Huitieme Edition.


The book has 1302 pages + a 36 pages supplement



From page 1130 to page 1259: Geometrie du triangle:


Good Luck!

1 σχόλιο:

  1. Suppose, $ H_a,H_b,H_c,H $ are the orthocenters of $ \Delta A"BC,\Delta B"CA,\Delta C"AB, \Delta ABC $.
    Note that, those nine-point centers are the midpoints of $ OH,OH_a,OH_b,OH_c $. So its enough to prove that $
    HH_aH_bH_c $ is cyclic.
    Now note that $ AHH_aA" $ is a parallelogram. So $ HH_a\parallel AA" $ and $ HH_a=AA" $ and similar for others.
    Suppose, $ A_2,B_2,C_2 $ are the diametrically opposite points of $ A",B",C" $ wrt $ (O) $. Note that $ AA_2,BB_2,CC_2 $ are concurrent at the reflection of $ D $ on $ O $.
    Now suppose, $ A_1,B_1,C_1 $ are the foot of the perpendiculars from $ O $ to $ AA_2,BB_2,CC_2 $. Note that $ OA_1\parallel AA" $ and $ OA_1=\frac {AA"}{2} $. Similar for others.
    So the configuration $ HH_aH_bH_c $v is homothetic to $ OA_1B_1C_1 $.
    Clearly $ O,A_1,B_1,C_1 $ lie on the circle with diameter $ OD' $ where $ D' $ is reflection of $ D $ on $ O $.
    So $ HH_aH_bH_c $ is cyclic. So done.

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