## 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.