Παρασκευή, 6 Δεκεμβρίου 2013

CONCURRENT EULER LINES

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

A1, Ab, Ac = the midpoints of A'P, A'B, A'C, resp.

B2, Bc, Ba = the midpoints of B'P, B'C, B'A, resp.

C3, Ca, Cb = the midpoints of C'P, C'A, C'B, resp.

For P = O the Euler lines of the triangles A1AbAc, B2BcBa, C3CaCb are concurrent (trivial case)

How about for P = I ??

In general, which is the locus of P such that the Euler lines of the triangles A1AbAc, B2BcBa, C3CaCb are concurrent?

Antreas P. Hatzipolakis, 5 Dec. 2013, Anopolis #1143

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The locus is a degree-7 circum-excentral-curve which passes through ETC-centers I, O and X(1138)=Isogonal conjugate of X(399)

1) For P=X(3)=O, point of concurrence is Z=O

2) For P=X(1138), Z=X(30)

3) For P=X(1)=I

Z = (2*a^4-2*(b+c)*a^3-(3*b^2+4*b*c+3*c^2)*a^2+2*(c^2-3*b*c+b^2)*(b+c)*a+(b^2-c^2)^2)/a : : (Trilinears)

= midpoint of: (1,442), (21,3649)

= on lines (1,442), (7,21), (30,551), (78,3826), (79,5426), (191,3338), (497,2475), (758,942), (950,3838), (958,3487), (962,4428), (1387,3636), (2646,5249), (3035,3812), (3651,5603), (3671,4640), (3897,5434)

= [2.022889134016317502091, 1.585022278218526006039, 1.609700227440945328741]

César Lozada, 6 Dec. 2013, Anopolis #1144

Παρασκευή, 29 Νοεμβρίου 2013

PERSPECTIVITY - LOCUS

Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P.

Denote:

Ab = the other than B intersection of the circle with diameter BA' and the circumcircle.

Ac = the other than C intersection of the circle with diameter CA' and the circumcircle.

Similarly (cyclically): Bc, Ba and Ca, Cb

Which is the locus of P such that:

1. ABC, Triangle bounded by (AbAc, BcBa, CaCb) are perspective?

Answer: The entire plane. Reference.

2. ABC, Triangle bounded by (BcCb, CaAc, AbBa) are perspective? The entire plane?

Special case: BcCb, CaAc, AbBa are concurrent.

3. Triangle bounded by (AbAc, BcBa, CaCb) and Triangle bounded by (BcCb, CaAc, AbBa) are perspective?

Antreas P. Hatzipolakis, 29 Nov. 2013

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Which is the locus of P such that:

1. ABC, Triangle bounded by (AbAc, BcBa, CaCb) are perspective?

Answer: The entire plane.

**** Perspector, if P=(u:v:w) (baricentric coordinates):

Q =( a^2 (a^2 - b^2 - c^2)/(-2 a^2 c^2 u^2 v^2 + 2 b^2 c^2 u^2 v^2 + 2 c^4 u^2 v^2 + a^4 u^2 v w - 2 a^2 b^2 u^2 v w + b^4 u^2 v w - 2 a^2 c^2 u^2 v w + 6 b^2 c^2 u^2 v w + c^4 u^2 v w - 2 a^2 b^2 u^2 w^2 + 2 b^4 u^2 w^2 + 2 b^2 c^2 u^2 w^2 + 2 a^4 v^2 w^2 - 2 a^2 b^2 v^2 w^2 - 2 a^2 c^2 v^2 w^2) : ... : ...)

Pairs of triangle centres {P, Q}: {X(69),X(64)}, {X(99),X(520)}

(Reference)

2. ABC, Triangle bounded by (BcCb, CaAc, AbBa) are perspective?

**** If P lies on Darboux cubic.

Special case: BcCb, CaAc, AbBa are concurrent.

**** If P lies on the sextic (No centers ETC):

-2 a^8 c^4 x^3 y^3 + 4 a^4 b^4 c^4 x^3 y^3 - 2 b^8 c^4 x^3 y^3 + 4 a^6 c^6 x^3 y^3 - 4 a^4 b^2 c^6 x^3 y^3 - 4 a^2 b^4 c^6 x^3 y^3 + 4 b^6 c^6 x^3 y^3 + 8 a^2 b^2 c^8 x^3 y^3 - 4 a^2 c^10 x^3 y^3 - 4 b^2 c^10 x^3 y^3 + 2 c^12 x^3 y^3 + a^10 c^2 x^3 y^2 z - 3 a^8 b^2 c^2 x^3 y^2 z + 2 a^6 b^4 c^2 x^3 y^2 z + 2 a^4 b^6 c^2 x^3 y^2 z - 3 a^2 b^8 c^2 x^3 y^2 z + b^10 c^2 x^3 y^2 z - 3 a^8 c^4 x^3 y^2 z + 8 a^6 b^2 c^4 x^3 y^2 z + 2 a^4 b^4 c^4 x^3 y^2 z - 7 b^8 c^4 x^3 y^2 z + 2 a^6 c^6 x^3 y^2 z - 6 a^4 b^2 c^6 x^3 y^2 z + 6 a^2 b^4 c^6 x^3 y^2 z + 14 b^6 c^6 x^3 y^2 z + 2 a^4 c^8 x^3 y^2 z - 10 b^4 c^8 x^3 y^2 z - 3 a^2 c^10 x^3 y^2 z + b^2 c^10 x^3 y^2 z + c^12 x^3 y^2 z + a^10 c^2 x^2 y^3 z - 3 a^8 b^2 c^2 x^2 y^3 z + 2 a^6 b^4 c^2 x^2 y^3 z + 2 a^4 b^6 c^2 x^2 y^3 z - 3 a^2 b^8 c^2 x^2 y^3 z + b^10 c^2 x^2 y^3 z - 7 a^8 c^4 x^2 y^3 z + 2 a^4 b^4 c^4 x^2 y^3 z + 8 a^2 b^6 c^4 x^2 y^3 z - 3 b^8 c^4 x^2 y^3 z + 14 a^6 c^6 x^2 y^3 z + 6 a^4 b^2 c^6 x^2 y^3 z - 6 a^2 b^4 c^6 x^2 y^3 z + 2 b^6 c^6 x^2 y^3 z - 10 a^4 c^8 x^2 y^3 z + 2 b^4 c^8 x^2 y^3 z + a^2 c^10 x^2 y^3 z - 3 b^2 c^10 x^2 y^3 z + c^12 x^2 y^3 z + a^10 b^2 x^3 y z^2 - 3 a^8 b^4 x^3 y z^2 + 2 a^6 b^6 x^3 y z^2 + 2 a^4 b^8 x^3 y z^2 - 3 a^2 b^10 x^3 y z^2 + b^12 x^3 y z^2 - 3 a^8 b^2 c^2 x^3 y z^2 + 8 a^6 b^4 c^2 x^3 y z^2 - 6 a^4 b^6 c^2 x^3 y z^2 + b^10 c^2 x^3 y z^2 + 2 a^6 b^2 c^4 x^3 y z^2 + 2 a^4 b^4 c^4 x^3 y z^2 + 6 a^2 b^6 c^4 x^3 y z^2 - 10 b^8 c^4 x^3 y z^2 + 2 a^4 b^2 c^6 x^3 y z^2 + 14 b^6 c^6 x^3 y z^2 - 3 a^2 b^2 c^8 x^3 y z^2 - 7 b^4 c^8 x^3 y z^2 + b^2 c^10 x^3 y z^2 + a^12 x^2 y^2 z^2 - 2 a^10 b^2 x^2 y^2 z^2 - a^8 b^4 x^2 y^2 z^2 + 4 a^6 b^6 x^2 y^2 z^2 - a^4 b^8 x^2 y^2 z^2 - 2 a^2 b^10 x^2 y^2 z^2 + b^12 x^2 y^2 z^2 - 2 a^10 c^2 x^2 y^2 z^2 + 6 a^8 b^2 c^2 x^2 y^2 z^2 - 4 a^6 b^4 c^2 x^2 y^2 z^2 - 4 a^4 b^6 c^2 x^2 y^2 z^2 + 6 a^2 b^8 c^2 x^2 y^2 z^2 - 2 b^10 c^2 x^2 y^2 z^2 - a^8 c^4 x^2 y^2 z^2 - 4 a^6 b^2 c^4 x^2 y^2 z^2 + 10 a^4 b^4 c^4 x^2 y^2 z^2 - 4 a^2 b^6 c^4 x^2 y^2 z^2 - b^8 c^4 x^2 y^2 z^2 + 4 a^6 c^6 x^2 y^2 z^2 - 4 a^4 b^2 c^6 x^2 y^2 z^2 - 4 a^2 b^4 c^6 x^2 y^2 z^2 + 4 b^6 c^6 x^2 y^2 z^2 - a^4 c^8 x^2 y^2 z^2 + 6 a^2 b^2 c^8 x^2 y^2 z^2 - b^4 c^8 x^2 y^2 z^2 - 2 a^2 c^10 x^2 y^2 z^2 - 2 b^2 c^10 x^2 y^2 z^2 + c^12 x^2 y^2 z^2 + a^12 x y^3 z^2 - 3 a^10 b^2 x y^3 z^2 + 2 a^8 b^4 x y^3 z^2 + 2 a^6 b^6 x y^3 z^2 - 3 a^4 b^8 x y^3 z^2 + a^2 b^10 x y^3 z^2 + a^10 c^2 x y^3 z^2 - 6 a^6 b^4 c^2 x y^3 z^2 + 8 a^4 b^6 c^2 x y^3 z^2 - 3 a^2 b^8 c^2 x y^3 z^2 - 10 a^8 c^4 x y^3 z^2 + 6 a^6 b^2 c^4 x y^3 z^2 + 2 a^4 b^4 c^4 x y^3 z^2 + 2 a^2 b^6 c^4 x y^3 z^2 + 14 a^6 c^6 x y^3 z^2 + 2 a^2 b^4 c^6 x y^3 z^2 - 7 a^4 c^8 x y^3 z^2 - 3 a^2 b^2 c^8 x y^3 z^2 + a^2 c^10 x y^3 z^2 - 2 a^8 b^4 x^3 z^3 + 4 a^6 b^6 x^3 z^3 - 4 a^2 b^10 x^3 z^3 + 2 b^12 x^3 z^3 - 4 a^4 b^6 c^2 x^3 z^3 + 8 a^2 b^8 c^2 x^3 z^3 - 4 b^10 c^2 x^3 z^3 + 4 a^4 b^4 c^4 x^3 z^3 - 4 a^2 b^6 c^4 x^3 z^3 + 4 b^6 c^6 x^3 z^3 - 2 b^4 c^8 x^3 z^3 + a^10 b^2 x^2 y z^3 - 7 a^8 b^4 x^2 y z^3 + 14 a^6 b^6 x^2 y z^3 - 10 a^4 b^8 x^2 y z^3 + a^2 b^10 x^2 y z^3 + b^12 x^2 y z^3 - 3 a^8 b^2 c^2 x^2 y z^3 + 6 a^4 b^6 c^2 x^2 y z^3 - 3 b^10 c^2 x^2 y z^3 + 2 a^6 b^2 c^4 x^2 y z^3 + 2 a^4 b^4 c^4 x^2 y z^3 - 6 a^2 b^6 c^4 x^2 y z^3 + 2 b^8 c^4 x^2 y z^3 + 2 a^4 b^2 c^6 x^2 y z^3 + 8 a^2 b^4 c^6 x^2 y z^3 + 2 b^6 c^6 x^2 y z^3 - 3 a^2 b^2 c^8 x^2 y z^3 - 3 b^4 c^8 x^2 y z^3 + b^2 c^10 x^2 y z^3 + a^12 x y^2 z^3 + a^10 b^2 x y^2 z^3 - 10 a^8 b^4 x y^2 z^3 + 14 a^6 b^6 x y^2 z^3 - 7 a^4 b^8 x y^2 z^3 + a^2 b^10 x y^2 z^3 - 3 a^10 c^2 x y^2 z^3 + 6 a^6 b^4 c^2 x y^2 z^3 - 3 a^2 b^8 c^2 x y^2 z^3 + 2 a^8 c^4 x y^2 z^3 - 6 a^6 b^2 c^4 x y^2 z^3 + 2 a^4 b^4 c^4 x y^2 z^3 + 2 a^2 b^6 c^4 x y^2 z^3 + 2 a^6 c^6 x y^2 z^3 + 8 a^4 b^2 c^6 x y^2 z^3 + 2 a^2 b^4 c^6 x y^2 z^3 - 3 a^4 c^8 x y^2 z^3 - 3 a^2 b^2 c^8 x y^2 z^3 + a^2 c^10 x y^2 z^3 + 2 a^12 y^3 z^3 - 4 a^10 b^2 y^3 z^3 + 4 a^6 b^6 y^3 z^3 - 2 a^4 b^8 y^3 z^3 - 4 a^10 c^2 y^3 z^3 + 8 a^8 b^2 c^2 y^3 z^3 - 4 a^6 b^4 c^2 y^3 z^3 - 4 a^6 b^2 c^4 y^3 z^3 + 4 a^4 b^4 c^4 y^3 z^3 + 4 a^6 c^6 y^3 z^3 - 2 a^4 c^8 y^3 z^3=0

3. Triangle bounded by (AbAc, BcBa, CaCb) and Triangle bounded by (BcCb, CaAc, AbBa) are perspective?

**** Always. The coordinates of perspector is very complicated.

First barycentric coordinate:

a^2 (32 a^14 b^2 c^8 u^7 v^5+96 a^12 b^4 c^8 u^7 v^5-480 a^10 b^6 c^8 u^7 v^5+352 a^8 b^8 c^8 u^7 v^5+352 a^6 b^10 c^8 u^7 v^5-480 a^4 b^12 c^8 u^7 v^5+96 a^2 b^14 c^8 u^7 v^5+32 b^16 c^8 u^7 v^5-96 a^12 b^2 c^10 u^7 v^5+64 a^10 b^4 c^10 u^7 v^5+608 a^8 b^6 c^10 u^7 v^5-1152 a^6 b^8 c^10 u^7 v^5+608 a^4 b^10 c^10 u^7 v^5+64 a^2 b^12 c^10 u^7 v^5-96 b^14 c^10 u^7 v^5+32 a^10 b^2 c^12 u^7 v^5-608 a^8 b^4 c^12 u^7 v^5+576 a^6 b^6 c^12 u^7 v^5+576 a^4 b^8 c^12 u^7 v^5-608 a^2 b^10 c^12 u^7 v^5+32 b^12 c^12 u^7 v^5+160 a^8 b^2 c^14 u^7 v^5+384 a^6 b^4 c^14 u^7 v^5-1088 a^4 b^6 c^14 u^7 v^5+384 a^2 b^8 c^14 u^7 v^5+160 b^10 c^14 u^7 v^5-160 a^6 b^2 c^16 u^7 v^5+416 a^4 b^4 c^16 u^7 v^5+416 a^2 b^6 c^16 u^7 v^5-160 b^8 c^16 u^7 v^5-32 a^4 b^2 c^18 u^7 v^5-448 a^2 b^4 c^18 u^7 v^5-32 b^6 c^18 u^7 v^5+96 a^2 b^2 c^20 u^7 v^5+96 b^4 c^20 u^7 v^5-32 b^2 c^22 u^7 v^5+64 a^14 b^2 c^8 u^6 v^6-128 a^12 b^4 c^8 u^6 v^6-64 a^10 b^6 c^8 u^6 v^6+256 a^8 b^8 c^8 u^6 v^6-64 a^6 b^10 c^8 u^6 v^6-128 a^4 b^12 c^8 u^6 v^6+64 a^2 b^14 c^8 u^6 v^6-128 a^12 b^2 c^10 u^6 v^6+384 a^10 b^4 c^10 u^6 v^6-256 a^8 b^6 c^10 u^6 v^6-256 a^6 b^8 c^10 u^6 v^6+384 a^4 b^10 c^10 u^6 v^6-128 a^2 b^12 c^10 u^6 v^6-64 a^10 b^2 c^12 u^6 v^6-256 a^8 b^4 c^12 u^6 v^6+640 a^6 b^6 c^12 u^6 v^6-256 a^4 b^8 c^12 u^6 v^6-64 a^2 b^10 c^12 u^6 v^6+256 a^8 b^2 c^14 u^6 v^6-256 a^6 b^4 c^14 u^6 v^6-256 a^4 b^6 c^14 u^6 v^6+256 a^2 b^8 c^14 u^6 v^6-64 a^6 b^2 c^16 u^6 v^6+384 a^4 b^4 c^16 u^6 v^6-64 a^2 b^6 c^16 u^6 v^6-128 a^4 b^2 c^18 u^6 v^6-128 a^2 b^4 c^18 u^6 v^6+64 a^2 b^2 c^20 u^6 v^6-32 a^16 b^2 c^6 u^7 v^4 w-32 a^14 b^4 c^6 u^7 v^4 w+480 a^12 b^6 c^6 u^7 v^4 w-800 a^10 b^8 c^6 u^7 v^4 w+160 a^8 b^10 c^6 u^7 v^4 w+672 a^6 b^12 c^6 u^7 v^4 w-608 a^4 b^14 c^6 u^7 v^4 w+160 a^2 b^16 c^6 u^7 v^4 w+128 a^14 b^2 c^8 u^7 v^4 w-96 a^12 b^4 c^8 u^7 v^4 w-1472 a^10 b^6 c^8 u^7 v^4 w+2912 a^8 b^8 c^8 u^7 v^4 w-1024 a^6 b^10 c^8 u^7 v^4 w-928 a^4 b^12 c^8 u^7 v^4 w+320 a^2 b^14 c^8 u^7 v^4 w+160 b^16 c^8 u^7 v^4 w-128 a^12 b^2 c^10 u^7 v^4 w+480 a^10 b^4 c^10 u^7 v^4 w-480 a^8 b^6 c^10 u^7 v^4 w-2112 a^6 b^8 c^10 u^7 v^4 w+3776 a^4 b^10 c^10 u^7 v^4 w-928 a^2 b^12 c^10 u^7 v^4 w-608 b^14 c^10 u^7 v^4 w-128 a^10 b^2 c^12 u^7 v^4 w-352 a^8 b^4 c^12 u^7 v^4 w+2944 a^6 b^6 c^12 u^7 v^4 w-2112 a^4 b^8 c^12 u^7 v^4 w-1024 a^2 b^10 c^12 u^7 v^4 w+672 b^12 c^12 u^7 v^4 w+320 a^8 b^2 c^14 u^7 v^4 w-352 a^6 b^4 c^14 u^7 v^4 w-480 a^4 b^6 c^14 u^7 v^4 w+2912 a^2 b^8 c^14 u^7 v^4 w+160 b^10 c^14 u^7 v^4 w-128 a^6 b^2 c^16 u^7 v^4 w+480 a^4 b^4 c^16 u^7 v^4 w-1472 a^2 b^6 c^16 u^7 v^4 w-800 b^8 c^16 u^7 v^4 w-128 a^4 b^2 c^18 u^7 v^4 w-96 a^2 b^4 c^18 u^7 v^4 w+480 b^6 c^18 u^7 v^4 w+128 a^2 b^2 c^20 u^7 v^4 w-32 b^4 c^20 u^7 v^4 w-32 b^2 c^22 u^7 v^4 w-8 a^18 c^6 u^6 v^5 w-56 a^16 b^2 c^6 u^6 v^5 w+384 a^14 b^4 c^6 u^6 v^5 w-704 a^12 b^6 c^6 u^6 v^5 w+304 a^10 b^8 c^6 u^6 v^5 w+528 a^8 b^10 c^6 u^6 v^5 w-704 a^6 b^12 c^6 u^6 v^5 w+256 a^4 b^14 c^6 u^6 v^5 w+24 a^2 b^16 c^6 u^6 v^5 w-24 b^18 c^6 u^6 v^5 w+40 a^16 c^8 u^6 v^5 w+288 a^14 b^2 c^8 u^6 v^5 w-672 a^12 b^4 c^8 u^6 v^5 w-416 a^10 b^6 c^8 u^6 v^5 w+1328 a^8 b^8 c^8 u^6 v^5 w-32 a^6 b^10 c^8 u^6 v^5 w-800 a^4 b^12 c^8 u^6 v^5 w+160 a^2 b^14 c^8 u^6 v^5 w+104 b^16 c^8 u^6 v^5 w-64 a^14 c^10 u^6 v^5 w-480 a^12 b^2 c^10 u^6 v^5 w-96 a^10 b^4 c^10 u^6 v^5 w+704 a^8 b^6 c^10 u^6 v^5 w+256 a^6 b^8 c^10 u^6 v^5 w+416 a^4 b^10 c^10 u^6 v^5 w-608 a^2 b^12 c^10 u^6 v^5 w-128 b^14 c^10 u^6 v^5 w+96 a^10 b^2 c^12 u^6 v^5 w+64 a^8 b^4 c^12 u^6 v^5 w-192 a^6 b^6 c^12 u^6 v^5 w-256 a^4 b^8 c^12 u^6 v^5 w+352 a^2 b^10 c^12 u^6 v^5 w-64 b^12 c^12 u^6 v^5 w+112 a^10 c^14 u^6 v^5 w+560 a^8 b^2 c^14 u^6 v^5 w+1216 a^6 b^4 c^14 u^6 v^5 w+1344 a^4 b^6 c^14 u^6 v^5 w+592 a^2 b^8 c^14 u^6 v^5 w+272 b^10 c^14 u^6 v^5 w-112 a^8 c^16 u^6 v^5 w-544 a^6 b^2 c^16 u^6 v^5 w-1056 a^4 b^4 c^16 u^6 v^5 w-672 a^2 b^6 c^16 u^6 v^5 w-176 b^8 c^16 u^6 v^5 w+32 a^4 b^2 c^18 u^6 v^5 w+32 a^2 b^4 c^18 u^6 v^5 w-64 b^6 c^18 u^6 v^5 w+64 a^4 c^20 u^6 v^5 w+160 a^2 b^2 c^20 u^6 v^5 w+128 b^4 c^20 u^6 v^5 w-40 a^2 c^22 u^6 v^5 w-56 b^2 c^22 u^6 v^5 w+8 c^24 u^6 v^5 w+64 a^16 b^2 c^6 u^5 v^6 w-192 a^14 b^4 c^6 u^5 v^6 w+64 a^12 b^6 c^6 u^5 v^6 w+320 a^10 b^8 c^6 u^5 v^6 w-320 a^8 b^10 c^6 u^5 v^6 w-64 a^6 b^12 c^6 u^5 v^6 w+192 a^4 b^14 c^6 u^5 v^6 w-64 a^2 b^16 c^6 u^5 v^6 w+192 a^14 b^2 c^8 u^5 v^6 w+128 a^12 b^4 c^8 u^5 v^6 w-1216 a^10 b^6 c^8 u^5 v^6 w+768 a^8 b^8 c^8 u^5 v^6 w+832 a^6 b^10 c^8 u^5 v^6 w-896 a^4 b^12 c^8 u^5 v^6 w+192 a^2 b^14 c^8 u^5 v^6 w-960 a^12 b^2 c^10 u^5 v^6 w+1216 a^10 b^4 c^10 u^5 v^6 w+1152 a^8 b^6 c^10 u^5 v^6 w-2176 a^6 b^8 c^10 u^5 v^6 w+832 a^4 b^10 c^10 u^5 v^6 w-64 a^2 b^12 c^10 u^5 v^6 w+704 a^10 b^2 c^12 u^5 v^6 w-2304 a^8 b^4 c^12 u^5 v^6 w+1152 a^6 b^6 c^12 u^5 v^6 w+768 a^4 b^8 c^12 u^5 v^6 w-320 a^2 b^10 c^12 u^5 v^6 w+704 a^8 b^2 c^14 u^5 v^6 w+1216 a^6 b^4 c^14 u^5 v^6 w-1216 a^4 b^6 c^14 u^5 v^6 w+320 a^2 b^8 c^14 u^5 v^6 w-960 a^6 b^2 c^16 u^5 v^6 w+128 a^4 b^4 c^16 u^5 v^6 w+64 a^2 b^6 c^16 u^5 v^6 w+192 a^4 b^2 c^18 u^5 v^6 w-192 a^2 b^4 c^18 u^5 v^6 w+64 a^2 b^2 c^20 u^5 v^6 w+8 a^18 b^2 c^4 u^7 v^3 w^2-24 a^16 b^4 c^4 u^7 v^3 w^2-32 a^14 b^6 c^4 u^7 v^3 w^2+224 a^12 b^8 c^4 u^7 v^3 w^2-336 a^10 b^10 c^4 u^7 v^3 w^2+112 a^8 b^12 c^4 u^7 v^3 w^2+224 a^6 b^14 c^4 u^7 v^3 w^2-288 a^4 b^16 c^4 u^7 v^3 w^2+136 a^2 b^18 c^4 u^7 v^3 w^2-24 b^20 c^4 u^7 v^3 w^2-40 a^16 b^2 c^6 u^7 v^3 w^2+96 a^14 b^4 c^6 u^7 v^3 w^2+576 a^12 b^6 c^6 u^7 v^3 w^2-1952 a^10 b^8 c^6 u^7 v^3 w^2+1584 a^8 b^10 c^6 u^7 v^3 w^2+544 a^6 b^12 c^6 u^7 v^3 w^2-1152 a^4 b^14 c^6 u^7 v^3 w^2+288 a^2 b^16 c^6 u^7 v^3 w^2+56 b^18 c^6 u^7 v^3 w^2+64 a^14 b^2 c^8 u^7 v^3 w^2-32 a^12 b^4 c^8 u^7 v^3 w^2-1088 a^10 b^6 c^8 u^7 v^3 w^2+2656 a^8 b^8 c^8 u^7 v^3 w^2-2368 a^6 b^10 c^8 u^7 v^3 w^2+1184 a^4 b^12 c^8 u^7 v^3 w^2-704 a^2 b^14 c^8 u^7 v^3 w^2+288 b^16 c^8 u^7 v^3 w^2-352 a^10 b^4 c^10 u^7 v^3 w^2+96 a^8 b^6 c^10 u^7 v^3 w^2+1088 a^6 b^8 c^10 u^7 v^3 w^2+1728 a^4 b^10 c^10 u^7 v^3 w^2-1248 a^2 b^12 c^10 u^7 v^3 w^2-1312 b^14 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a^6 b^16 c^2 u v^6 w^5-192 a^18 b^2 c^4 u v^6 w^5-128 a^16 b^4 c^4 u v^6 w^5+1216 a^14 b^6 c^4 u v^6 w^5-768 a^12 b^8 c^4 u v^6 w^5-832 a^10 b^10 c^4 u v^6 w^5+896 a^8 b^12 c^4 u v^6 w^5-192 a^6 b^14 c^4 u v^6 w^5+960 a^16 b^2 c^6 u v^6 w^5-1216 a^14 b^4 c^6 u v^6 w^5-1152 a^12 b^6 c^6 u v^6 w^5+2176 a^10 b^8 c^6 u v^6 w^5-832 a^8 b^10 c^6 u v^6 w^5+64 a^6 b^12 c^6 u v^6 w^5-704 a^14 b^2 c^8 u v^6 w^5+2304 a^12 b^4 c^8 u v^6 w^5-1152 a^10 b^6 c^8 u v^6 w^5-768 a^8 b^8 c^8 u v^6 w^5+320 a^6 b^10 c^8 u v^6 w^5-704 a^12 b^2 c^10 u v^6 w^5-1216 a^10 b^4 c^10 u v^6 w^5+1216 a^8 b^6 c^10 u v^6 w^5-320 a^6 b^8 c^10 u v^6 w^5+960 a^10 b^2 c^12 u v^6 w^5-128 a^8 b^4 c^12 u v^6 w^5-64 a^6 b^6 c^12 u v^6 w^5-192 a^8 b^2 c^14 u v^6 w^5+192 a^6 b^4 c^14 u v^6 w^5-64 a^6 b^2 c^16 u v^6 w^5+64 a^14 b^8 c^2 u^6 w^6-128 a^12 b^10 c^2 u^6 w^6-64 a^10 b^12 c^2 u^6 w^6+256 a^8 b^14 c^2 u^6 w^6-64 a^6 b^16 c^2 u^6 w^6-128 a^4 b^18 c^2 u^6 w^6+64 a^2 b^20 c^2 u^6 w^6-128 a^12 b^8 c^4 u^6 w^6+384 a^10 b^10 c^4 u^6 w^6-256 a^8 b^12 c^4 u^6 w^6-256 a^6 b^14 c^4 u^6 w^6+384 a^4 b^16 c^4 u^6 w^6-128 a^2 b^18 c^4 u^6 w^6-64 a^10 b^8 c^6 u^6 w^6-256 a^8 b^10 c^6 u^6 w^6+640 a^6 b^12 c^6 u^6 w^6-256 a^4 b^14 c^6 u^6 w^6-64 a^2 b^16 c^6 u^6 w^6+256 a^8 b^8 c^8 u^6 w^6-256 a^6 b^10 c^8 u^6 w^6-256 a^4 b^12 c^8 u^6 w^6+256 a^2 b^14 c^8 u^6 w^6-64 a^6 b^8 c^10 u^6 w^6+384 a^4 b^10 c^10 u^6 w^6-64 a^2 b^12 c^10 u^6 w^6-128 a^4 b^8 c^12 u^6 w^6-128 a^2 b^10 c^12 u^6 w^6+64 a^2 b^8 c^14 u^6 w^6+64 a^16 b^6 c^2 u^5 v w^6+192 a^14 b^8 c^2 u^5 v w^6-960 a^12 b^10 c^2 u^5 v w^6+704 a^10 b^12 c^2 u^5 v w^6+704 a^8 b^14 c^2 u^5 v w^6-960 a^6 b^16 c^2 u^5 v w^6+192 a^4 b^18 c^2 u^5 v w^6+64 a^2 b^20 c^2 u^5 v w^6-192 a^14 b^6 c^4 u^5 v w^6+128 a^12 b^8 c^4 u^5 v w^6+1216 a^10 b^10 c^4 u^5 v w^6-2304 a^8 b^12 c^4 u^5 v w^6+1216 a^6 b^14 c^4 u^5 v w^6+128 a^4 b^16 c^4 u^5 v w^6-192 a^2 b^18 c^4 u^5 v w^6+64 a^12 b^6 c^6 u^5 v w^6-1216 a^10 b^8 c^6 u^5 v w^6+1152 a^8 b^10 c^6 u^5 v w^6+1152 a^6 b^12 c^6 u^5 v w^6-1216 a^4 b^14 c^6 u^5 v w^6+64 a^2 b^16 c^6 u^5 v w^6+320 a^10 b^6 c^8 u^5 v w^6+768 a^8 b^8 c^8 u^5 v w^6-2176 a^6 b^10 c^8 u^5 v w^6+768 a^4 b^12 c^8 u^5 v w^6+320 a^2 b^14 c^8 u^5 v w^6-320 a^8 b^6 c^10 u^5 v w^6+832 a^6 b^8 c^10 u^5 v w^6+832 a^4 b^10 c^10 u^5 v w^6-320 a^2 b^12 c^10 u^5 v w^6-64 a^6 b^6 c^12 u^5 v w^6-896 a^4 b^8 c^12 u^5 v w^6-64 a^2 b^10 c^12 u^5 v w^6+192 a^4 b^6 c^14 u^5 v w^6+192 a^2 b^8 c^14 u^5 v w^6-64 a^2 b^6 c^16 u^5 v w^6+16 a^18 b^4 c^2 u^4 v^2 w^6+192 a^16 b^6 c^2 u^4 v^2 w^6-192 a^14 b^8 c^2 u^4 v^2 w^6-1216 a^12 b^10 c^2 u^4 v^2 w^6+2400 a^10 b^12 c^2 u^4 v^2 w^6-1216 a^8 b^14 c^2 u^4 v^2 w^6-192 a^6 b^16 c^2 u^4 v^2 w^6+192 a^4 b^18 c^2 u^4 v^2 w^6+16 a^2 b^20 c^2 u^4 v^2 w^6-64 a^16 b^4 c^4 u^4 v^2 w^6-320 a^14 b^6 c^4 u^4 v^2 w^6+1344 a^12 b^8 c^4 u^4 v^2 w^6-960 a^10 b^10 c^4 u^4 v^2 w^6-960 a^8 b^12 c^4 u^4 v^2 w^6+1344 a^6 b^14 c^4 u^4 v^2 w^6-320 a^4 b^16 c^4 u^4 v^2 w^6-64 a^2 b^18 c^4 u^4 v^2 w^6+64 a^14 b^4 c^6 u^4 v^2 w^6-448 a^12 b^6 c^6 u^4 v^2 w^6-1344 a^10 b^8 c^6 u^4 v^2 w^6+3456 a^8 b^10 c^6 u^4 v^2 w^6-1344 a^6 b^12 c^6 u^4 v^2 w^6-448 a^4 b^14 c^6 u^4 v^2 w^6+64 a^2 b^16 c^6 u^4 v^2 w^6+64 a^12 b^4 c^8 u^4 v^2 w^6+1088 a^10 b^6 c^8 u^4 v^2 w^6-1152 a^8 b^8 c^8 u^4 v^2 w^6-1152 a^6 b^10 c^8 u^4 v^2 w^6+1088 a^4 b^12 c^8 u^4 v^2 w^6+64 a^2 b^14 c^8 u^4 v^2 w^6-160 a^10 b^4 c^10 u^4 v^2 w^6-192 a^8 b^6 c^10 u^4 v^2 w^6+1984 a^6 b^8 c^10 u^4 v^2 w^6-192 a^4 b^10 c^10 u^4 v^2 w^6-160 a^2 b^12 c^10 u^4 v^2 w^6+64 a^8 b^4 c^12 u^4 v^2 w^6-704 a^6 b^6 c^12 u^4 v^2 w^6-704 a^4 b^8 c^12 u^4 v^2 w^6+64 a^2 b^10 c^12 u^4 v^2 w^6+64 a^6 b^4 c^14 u^4 v^2 w^6+448 a^4 b^6 c^14 u^4 v^2 w^6+64 a^2 b^8 c^14 u^4 v^2 w^6-64 a^4 b^4 c^16 u^4 v^2 w^6-64 a^2 b^6 c^16 u^4 v^2 w^6+16 a^2 b^4 c^18 u^4 v^2 w^6-16 a^20 b^2 c^2 u^2 v^4 w^6-192 a^18 b^4 c^2 u^2 v^4 w^6+192 a^16 b^6 c^2 u^2 v^4 w^6+1216 a^14 b^8 c^2 u^2 v^4 w^6-2400 a^12 b^10 c^2 u^2 v^4 w^6+1216 a^10 b^12 c^2 u^2 v^4 w^6+192 a^8 b^14 c^2 u^2 v^4 w^6-192 a^6 b^16 c^2 u^2 v^4 w^6-16 a^4 b^18 c^2 u^2 v^4 w^6+64 a^18 b^2 c^4 u^2 v^4 w^6+320 a^16 b^4 c^4 u^2 v^4 w^6-1344 a^14 b^6 c^4 u^2 v^4 w^6+960 a^12 b^8 c^4 u^2 v^4 w^6+960 a^10 b^10 c^4 u^2 v^4 w^6-1344 a^8 b^12 c^4 u^2 v^4 w^6+320 a^6 b^14 c^4 u^2 v^4 w^6+64 a^4 b^16 c^4 u^2 v^4 w^6-64 a^16 b^2 c^6 u^2 v^4 w^6+448 a^14 b^4 c^6 u^2 v^4 w^6+1344 a^12 b^6 c^6 u^2 v^4 w^6-3456 a^10 b^8 c^6 u^2 v^4 w^6+1344 a^8 b^10 c^6 u^2 v^4 w^6+448 a^6 b^12 c^6 u^2 v^4 w^6-64 a^4 b^14 c^6 u^2 v^4 w^6-64 a^14 b^2 c^8 u^2 v^4 w^6-1088 a^12 b^4 c^8 u^2 v^4 w^6+1152 a^10 b^6 c^8 u^2 v^4 w^6+1152 a^8 b^8 c^8 u^2 v^4 w^6-1088 a^6 b^10 c^8 u^2 v^4 w^6-64 a^4 b^12 c^8 u^2 v^4 w^6+160 a^12 b^2 c^10 u^2 v^4 w^6+192 a^10 b^4 c^10 u^2 v^4 w^6-1984 a^8 b^6 c^10 u^2 v^4 w^6+192 a^6 b^8 c^10 u^2 v^4 w^6+160 a^4 b^10 c^10 u^2 v^4 w^6-64 a^10 b^2 c^12 u^2 v^4 w^6+704 a^8 b^4 c^12 u^2 v^4 w^6+704 a^6 b^6 c^12 u^2 v^4 w^6-64 a^4 b^8 c^12 u^2 v^4 w^6-64 a^8 b^2 c^14 u^2 v^4 w^6-448 a^6 b^4 c^14 u^2 v^4 w^6-64 a^4 b^6 c^14 u^2 v^4 w^6+64 a^6 b^2 c^16 u^2 v^4 w^6+64 a^4 b^4 c^16 u^2 v^4 w^6-16 a^4 b^2 c^18 u^2 v^4 w^6-64 a^20 b^2 c^2 u v^5 w^6-192 a^18 b^4 c^2 u v^5 w^6+960 a^16 b^6 c^2 u v^5 w^6-704 a^14 b^8 c^2 u v^5 w^6-704 a^12 b^10 c^2 u v^5 w^6+960 a^10 b^12 c^2 u v^5 w^6-192 a^8 b^14 c^2 u v^5 w^6-64 a^6 b^16 c^2 u v^5 w^6+192 a^18 b^2 c^4 u v^5 w^6-128 a^16 b^4 c^4 u v^5 w^6-1216 a^14 b^6 c^4 u v^5 w^6+2304 a^12 b^8 c^4 u v^5 w^6-1216 a^10 b^10 c^4 u v^5 w^6-128 a^8 b^12 c^4 u v^5 w^6+192 a^6 b^14 c^4 u v^5 w^6-64 a^16 b^2 c^6 u v^5 w^6+1216 a^14 b^4 c^6 u v^5 w^6-1152 a^12 b^6 c^6 u v^5 w^6-1152 a^10 b^8 c^6 u v^5 w^6+1216 a^8 b^10 c^6 u v^5 w^6-64 a^6 b^12 c^6 u v^5 w^6-320 a^14 b^2 c^8 u v^5 w^6-768 a^12 b^4 c^8 u v^5 w^6+2176 a^10 b^6 c^8 u v^5 w^6-768 a^8 b^8 c^8 u v^5 w^6-320 a^6 b^10 c^8 u v^5 w^6+320 a^12 b^2 c^10 u v^5 w^6-832 a^10 b^4 c^10 u v^5 w^6-832 a^8 b^6 c^10 u v^5 w^6+320 a^6 b^8 c^10 u v^5 w^6+64 a^10 b^2 c^12 u v^5 w^6+896 a^8 b^4 c^12 u v^5 w^6+64 a^6 b^6 c^12 u v^5 w^6-192 a^8 b^2 c^14 u v^5 w^6-192 a^6 b^4 c^14 u v^5 w^6+64 a^6 b^2 c^16 u v^5 w^6-64 a^20 b^2 c^2 v^6 w^6+128 a^18 b^4 c^2 v^6 w^6+64 a^16 b^6 c^2 v^6 w^6-256 a^14 b^8 c^2 v^6 w^6+64 a^12 b^10 c^2 v^6 w^6+128 a^10 b^12 c^2 v^6 w^6-64 a^8 b^14 c^2 v^6 w^6+128 a^18 b^2 c^4 v^6 w^6-384 a^16 b^4 c^4 v^6 w^6+256 a^14 b^6 c^4 v^6 w^6+256 a^12 b^8 c^4 v^6 w^6-384 a^10 b^10 c^4 v^6 w^6+128 a^8 b^12 c^4 v^6 w^6+64 a^16 b^2 c^6 v^6 w^6+256 a^14 b^4 c^6 v^6 w^6-640 a^12 b^6 c^6 v^6 w^6+256 a^10 b^8 c^6 v^6 w^6+64 a^8 b^10 c^6 v^6 w^6-256 a^14 b^2 c^8 v^6 w^6+256 a^12 b^4 c^8 v^6 w^6+256 a^10 b^6 c^8 v^6 w^6-256 a^8 b^8 c^8 v^6 w^6+64 a^12 b^2 c^10 v^6 w^6-384 a^10 b^4 c^10 v^6 w^6+64 a^8 b^6 c^10 v^6 w^6+128 a^10 b^2 c^12 v^6 w^6+128 a^8 b^4 c^12 v^6 w^6-64 a^8 b^2 c^14 v^6 w^6)

Angel Montesdeoca, 30 Nov. 2013

ADDENDUM:

4. Triangle ABC, Triangle bounded by (BaCa, CbAb, AcBc) are perspective?

5. Triangle bounded by (BcCb, CaAc, AbBa),Triangle bounded by (BaCa, CbAb, AcBc) are perspective?

6. Triangle bounded by (AbAc, BcBa, CaCb),Triangle bounded by (BaCa, CbAb, AcBc) are perspective?

Antreas P. Hatzipolakis, 30 Nov. 2013


Σάββατο, 14 Σεπτεμβρίου 2013

PRIZE (Re: ORTHOCENTER - REFLECTIONS - CONCURRENT CIRCLES)

[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

For a proof I offer the book:

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

APH

Πέμπτη, 12 Σεπτεμβρίου 2013

CONCYCLIC

RE: [EGML] CONIC - LOCUS

Posted By: cesar_e_lozada

Fri Sep 6, 2013 10:28 pm

Dear Antreas:

They are concyclic for all points P.

The center of their circle is the complement of the isotomic of P.

Regards

Cιsar Lozada

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de Antreas Hatzipolakis

Enviado el: Jueves, 05 de Septiembre de 2013 06:28 p.m.

Para: anopolis@yahoogroups.com

Asunto: [EGML] CONIC - LOCUS

Let ABC be a triangle and A'B'C' the cevian triangle of P.

Denote:

Ab, Ac = the intersections of the circles with diameters BC, AA'

(near to B,C, resp.)

Bc, Ba = the intersections of the circles with diameters CA, BB'

Ca, Cb = the intersections of the circles with diameters AB, CC'

For P = H, the six points are concyclic.

For which P's are the six points lying on a conic?

APH

Anopolis #956


Πέμπτη, 5 Σεπτεμβρίου 2013

INTEGER SEQUENCES

In the book:

Marc Barbut, Bernard Monjardet: Ordre et Classification: Algebre et Combinatoire. Tome II. Hachette Universite, Paris, 1970.

are listed some integer sequences.

In pp. 44-46 the following:

1,3,19,219,4231,130023,6129859,.... It is A001035 in OEIS (The On-Line Encyclopedia of Integer Sequences)

1,4,29,355,6942,209527,9535241..... It is A000798 in OEIS

1,2,5,16,63,316,..... NOT FOUND IN OEIS

1,3,9,33,139,.... It is A001930 in OEIS 1,3,13,75,541,4683,47293,... It is A000670 in OEIS

In p. 101 the following:

1,2,5,,15,52,203,876,... It is A056273 in OEIS

0,1,3,7,15,31,63,.... It is A000225 in OEIS

0,0,1,6,25,90,301,... It is A000392 in OEIS

0,0,0,1,20,65,350,... NOT FOUND IN OEIS

0,0,0,0,1,15,140,... NOT FOUND IN OEIS

In pp. 165-166 the following:

1,4,18,166,7579,7828352,2414682040996,... It is A007153 in OEIS

1,3,8,28,208,.... NOT FOUND IN OEIS

1,2,4,12,81,... It is A001206 in OEIS

1,1,2,3,7,30,703,... NOT FOUND IN OEIS

BIBLIOGRAPHY:

For pages 1-78:

For pages 83-166:


Τετάρτη, 4 Σεπτεμβρίου 2013

PRIZE FOR CONCURRENT CIRCLES CONJECTURE

This "theorem" (conjecture) is still unproved (quoetd below).

Seiichi Kirikami has computed the coordinates for (P, P*) = (G, K)

Available here: Hyacinthos #21992

I offer the following books for proofs:

1. For an analytic proof by computing the homogeneous coordinates of the concurrence points:

RICHARD HEGER: ELEMENTE DER ANALYTISCHEN GEOMETRIE IN HOMOGENEN COORDINATEN (1872)

2. For a synthetic proof:

FRANK MORLEY & F. V. MORLEY: INVERSIVE GEOMETRY

3. For any other proof (by complex numbers, etc):

C. ZWIKKER: THE ADVANCED GEOMETRY OF PLANE CURVES AND THEIR APPLICATIONS.

Antreas

--- In Anopolis@yahoogroups.com, "Antreas" wrote:

>

> Let ABC be a triangle and P,P* two isogonal conjugate points.

> Denote: H1,H2,H3 = the orthocenters of PBC, PCA, PAB, resp. and

> Ha,Hb,Hc = the orthocenters of P*BC, P*CA, P*AB, resp.

>

> The circumcircles of:

> (1) H1HbHc, H2HcHa, H3HaHb

> (2) HaH2H3, HbH3H1, HcH1H2

>

> are concurrent.

>

> Figure: Here

>

> If P = (x:y:z), which are the points of concurrences?

>

> APH

>


Τετάρτη, 21 Αυγούστου 2013

Canarina canariensis

Re: RADICAL CENTERS - NPC - OI LINE

Posted By: amontes1949

Wed Aug 21, 2013 4:44 am

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

(Nab), (Nac) = the NPCs of AIB', AIC', resp.

(Nbc), (Nba) = the NPCs of BIC', BIA', resp.

(Nca), (Ncb) = the NPCs of CIA', CIB', resp.

R = the radical center of (Nbc), (Nca), (Nab)

S = the radical center of (Nba), (Ncb), (Nac)

*** (Trilinear ccordinates)

R = ((a + b - c) (a - b + c) (a^3 b - a b^3 + 2 a^3 c - 2 a b^2 c - b^3 c + a^2 c^2 - 3 a b c^2 - 3 b^2 c^2 - 2 a c^3 - 3 b c^3 - c^4) : (a - b - c) (a + b - c) (a^4 + 2 a^3 b - a^2 b^2 - 2 a b^3 + 3 a^3 c + 3 a^2 b c - b^3 c + 3 a^2 c^2 + 2 a b c^2 + a c^3 + b c^3) : (a - b - c) (a - b + c) (a^3 b + 3 a^2 b^2 + 3 a b^3 + b^4 + a^3 c + 2 a^2 b c + 3 a b^2 c + 2 b^3 c - b^2 c^2 - a c^3 - 2 b c^3)) T = ((a + b - c) (a - b + c) (2 a^3 b + a^2 b^2 - 2 a b^3 - b^4 + a^3 c - 3 a b^2 c - 3 b^3 c - 2 a b c^2 - 3 b^2 c^2 - a c^3 - b c^3): (a - b - c) (a + b - c) (a^3 b - a b^3 + a^3 c + 2 a^2 b c - 2 b^3 c + 3 a^2 c^2 + 3 a b c^2 - b^2 c^2 + 3 a c^3 + 2 b c^3 + c^4) : (a - b - c) (a - b + c) (a^4 + 3 a^3 b + 3 a^2 b^2 + a b^3 + 2 a^3 c + 3 a^2 b c + 2 a b^2 c + b^3 c - a^2 c^2 - 2 a c^3 - b c^3) )

R and S is a bicentric pair, then the midpoint of RS is a triangle center of trilinear ccordinates:

M = ( (a + b - c) (a - b + c) (3 a^3 b + a^2 b^2 - 3 a b^3 - b^4 + 3 a^3 c - 5 a b^2 c - 4 b^3 c + a^2 c^2 - 5 a b c^2 - 6 b^2 c^2 - 3 a c^3 - 4 b c^3 - c^4) : (a - b - c) (a + b - c) (a^4 + 3 a^3 b - a^2 b^2 - 3 a b^3 + 4 a^3 c + 5 a^2 b c - 3 b^3 c + 6 a^2 c^2 + 5 a b c^2 - b^2 c^2 + 4 a c^3 + 3 b c^3 + c^4) : (a - b - c) (a - b + c) (a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4 + 3 a^3 c + 5 a^2 b c + 5 a b^2 c + 3 b^3 c - a^2 c^2 - b^2 c^2 - 3 a c^3 - 3 b c^3) )

M lie on the central line X(1)X(3).

Ideal point of line RT is X(513).

This bicentric pair does not appear in the current edition of "BICENTRIC PAIRS OF POINTS" ( Clark Kimberling)

I suggest as a flower name for this bicentric pair: Canarina

Angel Montesdeoca

Anopolis #860

Κυριακή, 11 Αυγούστου 2013


Postcard sent in 1897 from Salonica, Greece, then Ottoman Empire.

The text is cryptographed (encrypted)

Σάββατο, 20 Ιουλίου 2013

ORTHOPOLE CIRCLES

Let (a,b,c,d) be a complete quadrilateral.

Denote:

A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) = the antimedial triangles of (b,c,d), (c,d,a),(d,a,b), (a,b,c), resp.

Oa, Ob, Oc, Od = the circumcenters of A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c), resp. [=orthocenters of (b,c,d), (c,d,a),(d,a,b), (a,b,c), lying on the Steiner line]

M = the Miquel point of the quadrilateral (= the point of concurrence of the circumcircles of (b,c,d), (c,d,a),(d,a,b), (a,b,c))

Are the orthopoles of MOa, MOb, Moc, MOd with respect A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) concyclic?

The orthopoles lie on the NPCs of A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) [ = circumcircles of (b,c,d), (c,d,a),(d,a,b), (a,b,c)]

Antreas P. Hatzipolakis, 19 July 2013

Παρασκευή, 5 Ιουλίου 2013

CONCURRENT CIRCLES - ORTHOCENTERS - ISOGONAL CONJUGATE POINTS


Let ABC be a triangle and P,P* two isogonal conjugate points. Denote: H1,H2,H3 = the orthocenters of PBC, PCA, PAB, resp. and Ha,Hb,Hc = the orthocenters of P*BC, P*CA, P*AB, resp.


The circumcircles of: (1) H1HbHc, H2HcHa, H3HaHb (2) HaH2H3, HbH3H1, HcH1H2 are concurrent.
Antreas P. Hatzipolakis, 5 July 2013

Κυριακή, 30 Ιουνίου 2013

ANOPOLIS CIRCLE (2)

Let ABC be a triangle and L1,L2,L3 the Euler lines of IBC,ICA,IAB, resp. (concurrent at Schiffler point).
Denote:
E1 = the point of concurrence of the reflections of L1 in the sidelines of IBC (on the circumcircle of IBC)
E2 = the point of concurrence of the reflections of L2 in the sidelines of ICA (on the circumcircle of ICA)
E3 = the point of concurrence of the reflections of L3 in the sidelines of IAB (on the circumcircle of IAB)

The points E1,E2,E3 lie on the circle with diameter IO.
See also ANOPOLIS CIRCLE (1)
Antreas P. Hatzipolakis, 30 June 2013

Triangle Construction h_a, m_a, |b-c|/a

To construct triabgle ABC if are given:

altitude h_a, median m_a and the ratio |b-c|/a = m/n

Reference:

EUCLID, June 1969, p. 420-1

Παρασκευή, 7 Ιουνίου 2013

CONCURRENT EULER LINES

Let ABC be a triangle, IAb, IAc the trisectors of BIC (with Ab, Ac near B, C, resp.), IBc, IBa the trisectors of CIA (with Bc, Ba near C, A, resp.) and ICa, ICb the trisectors of AIB (with Ca, Cb near A, B, resp.)

Conjecture:

The Euler Lines of IBcCb, ICaAc, IAbBa are concurrent.

Antreas P. Hatzipolakis, 7 June 2013

***************************************

The Euler lines do not concur.

However, your configuration does provide a construction for a new ellipse through Ab, Ac, Bc, Ba, Ca, Cb, with some new (and one existing) centers:

Center of ellipse = (non-ETC search 1.883071694412417), on line 1,3604.

Let A' be the intersection of tangents to ellipse at Ab and Ac, and define B', C' cyclically.

Let A" be the intersection of tangents to ellipse at Ba and Ca, and define A", B" cyclically.

Let A* be the intersection of tangents to ellipse at Bc and Cb, and define B*, C* cyclically. The lines AA', BB', CC' concur in P1 = (non-ETC search 1.178825206518684), the perspector of the ellipse.

The lines AA"A*, BB"B*, CC"C* concur in X(3604).

The lines A'A", B'B", C'C" concur in P2 = (non-ETC search 1.249947393611434).

The lines A'A*, B'B*, C'C* concur in P3 = (non-ETC search 1.043484806530543).

P2 and P3 are collinear with X(3604).

Some variations to explore:

1) other starting points besides I

2) substitute external angle trisectors for internal

Randy Hutson Anopolis #377

Τρίτη, 4 Ιουνίου 2013

Κυριακή, 2 Ιουνίου 2013

N1N2N3 - PERSPECTIVE

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral tr.), N1,N2,N3 the NPC centers of IBC, ICA, IAB, resp. and Na,Nb,Nc the NPC centers of A'BC, B'CA, C'AB, resp.

1. The triangles A'B'C', N1N2N3 are perspective.

Perspector:

(aabc+a(a+b+c)(bb+4bc+cc-aa) , ... , ...)

Barry Wolk, Anopolis #347

ETC X(5506)

2. The triangles N1N2N3, NaNbNc are perspective (?).

Antreas P. Hatzipolakis, 2 June 2013

CONCURRENT CIRCUMCIRCLES - N1N2N3, NaNbNc

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral triangle) N1,N2,N3 the NPC centers of IBC, ICA, IAB, resp. and Na, Nb, Nc the NPC centers of A'BC, B'CA, C'AB, resp.

1. The circumcircles of ABC, AN2N3, BN3N1, CN1N2 are concurrent.

********************

It is now the center X(5606) in ETC

*******************

2. The circumcircles of ABC, ANbNc, BNcNa, CNaNb are concurent.

Note: The circumcenter of NaNbNc is the O.

Antreas P. Hatzipolakis, 2 June 2013

Addendum (26 - 11 - 2013)

1'. The circumcircles of N1BC, N2CA, N3AB are concurrent.

2'. The circumcircles of NaBC, NbCA, NcAB are concurrent.

APH, Anopolis #1119

*******************************

*** 1'. X(502)

*** 2'. ( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... )

with (6-9-13)-search number 2.5719845710987353841258271936

Angel Montesdeoca, Anopolis #1120

*******************************

( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... ) = R X[65]-(2r+R) X[1365], is on lines {{1,149},{10,1109},{36,759},{37,115},{65,1365},{162,1838},{267,3336},{897,1738},{1054,1247},{1737,2166},{2218,2915}}.

isogonal conjugate X(5127)

X(2245) cross-conjugate of X(226)

trilinear pole of line X(661) X(2294)

X(3) isoconjugate of X(2074)

X(21) isoconjugate of X(5172)

trilinear product of X(523) & X(1290)

barycentric product of X(1290) & X(1577)

antigonal of X(502)

Peter J.C. Moses, 2 Dec 2013

***************

It is now X(5620) = ISOGONAL CONJUGATE OF X(5127) in ETC

Τετάρτη, 29 Μαΐου 2013

A NPC CENTER ON THE EULER LINE

Let ABC be a triangle and N1,N2,N3 the NPC centers of OBC, OCA, OAB, resp.

The NPC center of N1N2N3 lies on the Euler line of ABC.

Antreas P. Hatzipolakis, 30 May 2013

***************************************************** The barycentric coordinates of the NPC center of N1N2N3 are:

( 2a^10 - 5a^8(b^2+c^2) + 2a^6(b^4+5b^2c^2+c^4) + a^4(4b^6-5b^4c^2-5b^2c^4+4c^6) - a^2(b^2-c^2)^2(4b^4+5b^2c^2+4c^4) + (b^2-c^2)^4(b^2+c^2) : ... : ...)

with (6-9-13)-search number: 4.7800096839999025703058

Angel Montesdeoca, Anopolis #331

CONCURRENT CIRCLES

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

L11 = the perpendicular line to AA' at A'

L22 = the perpendicular line to BB' at B'

L33 = the perpendicular line to CC' at C'

(ie the lines L11,L22,L33 bound the antipedal triangle of I wrt A'B'C')

L12 = the reflection of L11 in BB'

L13 = the reflection of L11 in CC'

M12 = the parallel to L12 through B'

M13 = the parallel to L13 through C'

A" = line M12 /\ line M13

Similarly B" and C".

O1 = the circumcenter of the triangle A"B'C'

O2 = the circumcenter of the triangle B"C'A'

O3 = the circumcenter of the triangle C"A'B'

Conjecture 1:

The points I, O1, O2, O3 are concyclic. Center of the circle?

Conjecture 2: The circumcircles (O1), (O2), (O3) are concurrent. Point?

Antreas P. Hatzipolakis, 29 May 2013

********************************************************

The points I, O1, O2, O3 are concyclic. Center of the circle (barycentrics)

( a(b+c)(a^5-2a^3(b^2+c^2) + a(b^4-b^2c^2+c^4) - a^2b c(b+c)+ b(b-c)^2c(b+c) ) : ... : ... ).

The circumcircles (O1), (O2), (O3) are concurrent.

Point of concurrence:

( a(a^6-a^5(b+c) - a^4(b+c)^2 + (b^2-c^2)^2(b^2+c^2) + a^3(2b^3+b^2c+b c^2+2c^3) - a^2(b^4-b^3c-3b^2c^2-b c^3+c^4) - a(b^5+b^4c+b c^4+c^5)) : ... : ... )

Figure

Angel Montedeoca, Anopolis #327

Points X(5496), X(5497) in ETC

Τρίτη, 28 Μαΐου 2013

CONCYCLIC CIRCUMCENTERS

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

L11 = the perpendicular line to AA' at A'

L22 = the perpendicular line to BB' at B'

L33 = the perpendicular line to CC' at C'

(ie the lines L11,L22,L33 bound the antipedal triangle of I wrt A'B'C')

L12 = the reflection of L11 in BB'

L13 = the reflection of L11 in CC'

L21 = the reflection of L22 in AA'

L23 = the reflection of L22 in CC'

L31 = the reflection of L33 in AA'

L32 = the reflection of L33 in BB'

O1 = the circumcenter of the triangle bounded by the lines (L11,L12,L13)

O2 = the circumcenter of the triangle bounded by the lines (L21,L22,L23)

O3 = the circumcenter of the triangle bounded by the lines (L31,L32,L33)

Conjecture 1:

O1, O2, O3 and O [circumcenter of ABC] are concyclic.

Denote:

Oa = the circumcenter of the triangle bounded by the lines (L11,L21,L31)

Ob = the circumcenter of the triangle bounded by the lines (L12,L22,L32)

Oc = the circumcenter of the triangle bounded by the lines (L13,L23,L33)

Conjecture 2:

The circumcenter of the triangle OaObOc is the I.

Antreas P. Hatzipolakis, 28 May 2013

*******************************************************

Conjecture 1:

Yes, O1, O2, O3 and O are concyclic. The center of the circle is:

( a^2 ( a^7 (b + c) - a^6 (b^2 + c^2) - a^5 (3 b^3 + 2 b^2 c + 2 b c^2 + 3 c^3) + a^4 (3 b^4 - b^3 c + 4 b^2 c^2 - b c^3 + 3 c^4) + a^3 (3 b^5 + b^4c + 2 b^3c^2 + 2 b^2 c^3 + b c^4 + 3 c^5) - a^2 (3 b^6 - 2 b^5 c - 2 b c^5 + 3 c^6) - a (b^7 - b^4 c^3 - b^3 c^4 + c^7) + (b^2-c^2)^2(b^4 - b^3 c - b^2 c^2 - b c^3 + c^4)):... :...),

with (6-9-13)-search number: 0.025111873257385778374122883

Angel Montesdeoca, Anopolis #321

Point X(5495) in ETC

Σάββατο, 25 Μαΐου 2013

EULER LINES

NOTATIONS:

Let La, Lb, Lc be three lines, A,B,C three respective points and 1,2,3 three respective lines.

Denote:

rLa, rLb, rLc = the reflections of La, Lb, Lc in 1,2,3, resp.

pLa, pLb, pLc = the parallels to La, Lb, Lc through A, B, C, resp.

CONCURRENT EULER LINES.

Let ABC be a triangle and A'B'C' the cevian triangle I.

Denote:

Ab,Ac: = the reflections of A' in BB', CC', resp.

Bc,Ba: = the reflections of B' in CC', AA', resp.

Ca,Cb: = the reflections of C' in AA', BB', resp.

1,2,3: = the cevians AA', BB', CC' of I (bisectors)

1. L, La,Lb,Lc: = the Euler lines of ABC, AAbAc, BBcBa, CCaCb, resp.

1.1. La,Lb,Lc concur at the infinity point of L = X30 (ie L, La,Lb,Lc are parallel)

1.2. rLa, rLb, rLc are concurrent.

1.3. prLa, prLb, prLc are concurrent at a point on the circumcircle. The point is the antipode of the Euler line reflection point (ie the point of concurrence of the reflections of the Euler in the sidelines of ABC) = the isogonal conjugate of the infinity point of the Euler line (X30) = X74.

2. Da, Db, Dc: = the Euler lines of the triangles A'AbAc, B'BcBa, C'CaCb, resp.

2.1. Da, Db, Dc are concurrent at I. (I is the common circumcenter of the triangles)

2.2. pDa, pDb, pDc are concurrent.

Antreas P. Hatzipolakis, 24-25 May 2013

******************************************

1.2

The lines intersect at:

( a(a^9 - a^8(b+c) - a^7(b-c)^2 + a^6(2b^3-b^2c-b*c^2+2c^3) - a^5(3b^4+b^3c-7b^2c^2+b*c^3+3c^4) + 4a^4b*c(b-c)^2(b+c) + a^3(b^2-c^2)^2(5b^2-4b*c+5c^2) - a^2(b-c)^2(2b^5+5b^4c+b^3c^2+b^2c^3+5b*c^4+2c^5) - a(b^2-c^2)^2(2b^4-3b^3c+5b^2c^2-3b*c^3+2c^4) + (b-c)^4(b+c)^3(b^2+c^2)) : ... : ...),

with (6-9-13)-search number: 5.63864638926896001044233914

Angel Montesdeoca Anopolis, #301

On lines {{1,2779},{21,104},{36,1725},{65,74},{125,860}}

(2 r + R) X[110] - 4 (r + R) X[1385]

2 R X[65] + (2 r + R) X[74]

Peter Moses 27 May 2013

2.2

The lines are concurrent in X(80), reflection of incenter in Feuerbach point.

Angel Montesdeoca Anopolis, #308

Δευτέρα, 20 Μαΐου 2013

EXCENTERS - CIRCUMCENTERS - CONICS

Let ABC be a triangle and P a point.

Denote:

Iab, Iac = the excenters of IBC respective to angles PBC, PCB.

Ibc, Iba = the excenters of ICA respective to angles PCA, PAC.

Ica, Icb = the excenters of IAB respective to angles PAB, PBA.

For P = O, the six excenters are concyclic, lying on the circumcircle (O).

Are they always lying on a conic? And for which points P the conic is circle?

Denote:

Oa, O'a = the circumcenters of PIbcIcb, PIbaIca, resp.

Ob, O'b = the circumcenters of PIcaIac, PIcbIab, resp.

Oc, O'c = the circumcenters of PIabIba, PIacIbc, resp.

The triangles OaObOc, O'aO'bO'c are perspective.

The line segments OaO'a, ObO'b, OcO'c are bisected by the perspector P' of the triangles, therefore the six points lie on a conic with center P'.

For which points P the conic is circle?

Antreas P. Hatzipolakis, 20 May 2013

Κυριακή, 19 Μαΐου 2013

ANOPOLIS CIRCLE

Let ABC be a triangle.

Denote:

(N1),(N2), (N3) = the NPCs of IBC, ICA, IAB, resp.

(12), (13) = the reflections of (N1) in BI, CI, resp.

(23), (21) = the reflections of (N2) in CI, AI, resp.

(31), (32) = the reflections of (N3) in AI, B1, resp.

The six centers 12,13,23,21,31,32 are concyclic.

The circle has diameter IO.

Perspectivity:

The circles (I), (21), (31) concur at a point A'.

The circles (I), (32), (12) concur at a point B'.

The circles (I), (13), (23) concur at a point C'.

The triangles ABC, A'B'C' are perspective (??)

Antreas P. Hatzipolakis, 19 May 2013

********************************************

Midpoint of OI = X(1385)

12 = {a (a^2-b^2+a c-2 b c-c^2),-a^3+a b^2+a^2 c+2 a b c-2 b^2 c+a c^2-3 b c^2-c^3,-c^2 (b+c)} 13 = {a (a^2+a b-b^2-2 b c-c^2),-b^2 (b+c),-a^3+a^2 b+a b^2-b^3+2 a b c-3 b^2 c+a c^2-2 b c^2}

ETC points on the Anopolis circle {1,3,1083,3109,3110}

The orthogonal projection of I on a line through O is also on the circle .. hence X(3109) & X(3110). So too X(1083) as it is the orthogonal projection of I on line X(3) X(667)

Also the orthogonal projection of O on a line through I is also on the circle

A non ETC point on the Anopolis circle

a^2 (a^4 b^2-a^2 b^4-2 a^4 b c+a b^4 c+a^4 c^2+a^2 b^2 c^2-b^3 c^3-a^2 c^4+a b c^4) ::

on lines {{1,667},{3,238},{35,1083},{41,813}}

A’ = reflection of midpoint of AI in 21 31 = {(a-b-c) (b-c)^2 (a+b-c) (a-b+c),-(a-b)^2 b^2 (a+b-c),-(a-c)^2 c^2 (a-b+c)}

A’B’C’ is perspective to ABC at X(59)

Also to the tangential triangle at:

a^2 (a^9-3 a^8 b+8 a^6 b^3-6 a^5 b^4-6 a^4 b^5+8 a^3 b^6-3 a b^8+b^9-3 a^8 c+12 a^7 b c-12 a^6 b^2 c-12 a^5 b^3 c+30 a^4 b^4 c-12 a^3 b^5 c-12 a^2 b^6 c+12 a b^7 c-3 b^8 c-12 a^6 b c^2+33 a^5 b^2 c^2-21 a^4 b^3 c^2-21 a^3 b^4 c^2+33 a^2 b^5 c^2-14 a b^6 c^2+2 b^7 c^2+8 a^6 c^3-12 a^5 b c^3-21 a^4 b^2 c^3+48 a^3 b^3 c^3-21 a^2 b^4 c^3-4 a b^5 c^3+2 b^6 c^3-6 a^5 c^4+30 a^4 b c^4-21 a^3 b^2 c^4-21 a^2 b^3 c^4+18 a b^4 c^4-2 b^5 c^4-6 a^4 c^5-12 a^3 b c^5+33 a^2 b^2 c^5-4 a b^3 c^5-2 b^4 c^5+8 a^3 c^6-12 a^2 b c^6-14 a b^2 c^6+2 b^3 c^6+12 a b c^7+2 b^2 c^7-3 a c^8-3 b c^8+c^9)::

= (2r - R) (r^2 + 6 r R + 8 R^2 - s^2) X[1486] – 4 r (r^2 + 5 r R + 4 R^2 - s^2) X[1618]

On line {1486, 1618}

Search = 0.89807482985690351940

Peter J. C. Moses, 19 May 2013

Σάββατο, 18 Μαΐου 2013

COLLINEARITY

Let ABC be a triangle and P, Q two points.

Denote:

PQa = the isogonal conjugate of P wrt QBC

PQb = the isogonal conjugate of P wrt QCA

PQc = the isogonal conjugate of P wrt QAB

QPa = the isogonal conjugate of Q wrt PBC

QPb = the isogonal conjugate of Q wrt PCA

QPc = the isogonal conjugate of Q wrt PAB

Apq = PQaQPa /\ BC

Bpq = PQbQPb /\ CA

Cpq = PQcQPc /\ AB

Conjecture: The points Apq, Bpq, Cpq are collinear.

Let R be another point. We have the lines (if the conjecture is true):

ApqBpqCpq, AqrBqrCqr, ArpBrpCrp.

Are they concurrent? Do they bound a triangle in perspective with ABC?

Antreas P. Hatzipolakis, 18 May 2013

Παρασκευή, 17 Μαΐου 2013

ORTHOCENTERS

Let ABC be a triangle and A'B'C' the cevian triangle of P = I.

Denote:

Ab,Ac = the reflections of A' in BB', CC', resp.

Bc,Ba = the reflections of B' in CC', AA', resp.

Ca,Cb = the reflections of C' in AA', BB', resp.

Ha, Hb, Hc = the orthocenters of the triangles A'AbAc, B'BcCa, C'CaCb, resp.

H'a, H'b, H'c = the reflections of Ha, Hb, Hc in AA', BB', CC', resp.

Conjecture 1.:

The triangles ABC, HaHbHc are perspective.

Conjecture 2.:

The points H'a, H'b, H'c are collinear

Locus of variable P such that

1. ABC, HaHbHc are perspective

2. H'a, H'b, H'c are collinear?

Antreas P. Hatzipolakis, 17 May 2013

CONCURRENT CIRCUMCIRCLES - CONCYCLIC POINTS

Let ABC be a triangle, Na,Nb,Nc the NPC centers of IBC, ICA, IAB, resp. and Oa, Ob, Oc the circumcenters of NaBC, NbCA, NcAB, resp.

1. The Euler line of NaNbNc is the line INF of ABC (O of NaNbNc = N of ABC, H of NaNbNc = I of ABC, F of ABC = ?? of NaNbNc).

2. The circumcircles (Oa),(Ob),(Oc) of NaBC, NbCA, NcAB, resp. concur at a point Q on the circumcircle of NaNbNc.

Which point is the Q wrt 1. ABC 2.NaNbNc ?

Antreas P. Hatzipolakis, 17 May 2013

Πέμπτη, 16 Μαΐου 2013

RADICAL CENTER- EULER LINE

Let ABC be a triangle and A'B'C' the cevian triangle of P = H (orthic tr.)

Denote:

Ab,Ac = the reflections of A' in BB', CC', resp.

Bc,Ba = the reflections of B' in CC', AA', resp.

Ca,Cb = the reflections of C' in AA', BB', resp.

Na, Nb, Nc = The NPC centers of A'AbAc, B'BcBa, C'CaCb, resp.

O* = The circumcenter of NaNbNc. It is the NPC center of A'B'C'.

R* = the radical center of (Na), (Nb), (Nc)

The NPC circles (Na), (Nb), (Nc) concur on the NPC of A'B'C'. The point of concurrence, the R*, is the Poncelet point of H wrt A'B'C' and since the H of ABC is the I of A'B'C', the point is the Feuerbach point [1] of A'B'C' = the center of the Feuerbach hyperbola of A'B'C'.

The points R*, H, O* are collinear. The line is the INF-line of A'B'C'

Generalization:

Let P = point on the Euler line of ABC.

Conjecture:

The points R*, P, O* are collinear.

Locus:

P = a variable point.

Which is the locus of P such that R*,P,O* are collinear?

Euler Line + ???

Is the incenter I on the locus?

Note [1]:

The Feuerbach point of ABC:

Let ABC be a triangle and A'B'C' the cevian triangle of I. Denote:

Ab, Ac = the reflections of A in BB', CC', resp.

Bc, Ba = the reflectuons of B in CC', AA', resp.

Ca, Cb = the reflections of C in AA', BB', resp.

The NPCs of AAbAc, BBcBa, CCaCb (and ABC) concur at Feuerbach point of ABC.

Antreas P. Hatzipolakis, 16 May 2013

Τετάρτη, 15 Μαΐου 2013

NPC center of the Cevian triangle of I

Let ABC be a triangle, A'B'C' the cevian triangle of I, I' the isogonal conjugate of I wrt A'B'C' and N' the NPC center of A'B'C'.

Conjecture 1:

The points I, N',I' are collinear.

Conjecture 2:

Let A"B"C" be the pedal triangle of I' wrt A'B'C'.

I' is the NPC center A"B"C".(Randy Hutson)

The line II' is the Euler line of A"B"C"

Antreas P. Hatzipolakis, 15 May 2013

Anopolis #255 (Re: NPC center of the cevian triangle of I)

 

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