HYACINTHOS 28494

 

[Antreas P. Hatzipolakis]:

 

 

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

Denote:

(Na), (Nb), (Nc) = the NPCs of PBC, PCA, PAB, resp.

D = the Poncelet point of ABCP.

A", B", C" = antipodes of A', B', C' in (Na), (Nb), (Nc), resp.

 

The perpendicular from A" to DP intersets again (Na) at A*
The perpendicular from B" to DP intersets again (Nb) at B*

The perpendicular from C" to DP intersets again (Nc) at C*  

A*, B*, C*, D are concyclic.
Center of the circle?


[César Lozada]:

 

> A*, B*, C*, D are concyclic. Center of the circle?

 

Seems to be true. Unfortunately, expressions are very long and it is difficult to get a general expression for center.

 

Also, if

The perpendicular from A’ to DP intersets again (Na) at A**
The perpendicular from B’ to DP intersets again (Nb) at B**

The perpendicular from C’ to DP intersets again (Nc) at C**
then

A**, B**, C**, D seems also to be concyclic

Again, we have very long expressions.

 

Examples:

Let O*, O** be the centers of both circles, resp.

 

O*( X(1) ) = midpoint of X(80) and X(15343)

= (b+c)*a^6-8*a^5*b*c+6*b*c*(b+c)*a^4+(b^4+c^4-10*b^2*c^2)*a^3-(b+c)*(2*c^4+3*b*c^3-11*b^2*c^2+3*b^3*c+2*b^4)*a^2-(b^2-c^2)^2*a*(b^2-8*b*c+c^2)+(b^2-c^2)^2*(b+c)*(c^2-3*b*c+b^2) : : (barys)

= 3*X(5587)-X(18341)

= on lines: {1, 10774}, {10, 900}, {11, 23869}, {12, 13604}, {80, 3120}, {214, 11814}, {355, 18342}, {546, 946}, {2802, 21093}, {5587, 18341}, {6788, 24188}

= midpoint of X(i) and X(j) for these {i,j}: {80, 15343}, {355, 18342}

= [ 0.5560922401364657, 2.0808753187114790, 1.9434005350442170 ]

 

O**( X(1) ) = midpoint of X(1) and X(18341)

= (b+c)*a^9+(b^2-10*b*c+c^2)*a^8-(b+c)*(5*c^2-16*b*c+5*b^2)*a^7-2*(c^2-6*b*c+b^2)*(c^2-3*b*c+b^2)*a^6+(b+c)*(9*c^4-37*b*c^3+57*b^2*c^2-37*b^3*c+9*b^4)*a^5-b*c*(5*b^2-23*b*c+5*c^2)*(b-c)^2*a^4-(b^2-c^2)*(b-c)*a^3*(7*c^4-12*b*c^3+23*b^2*c^2-12*b^3*c+7*b^4)+(b^2-c^2)^2*a^2*(2*b^4-4*b^3*c+7*b^2*c^2-4*b*c^3+2*c^4)+(b^2-c^2)^3*(b-c)*a*(2*c^2-b*c+2*b^2)+(b^2-c^2)^4*(-b^2+b*c-c^2) : : (barys)

= 3*X(5886)-X(18342)

= on lines: {1, 18341}, {10, 140}, {496, 11798}, {900, 946}, {5533, 15529}, {5886, 18342}

= midpoint of X(1) and X(18341)

= [ 0.5496512431333436, -2.7170886604654480, 5.2680406730912210 ]

 

O*(X(4)) = O**(X(4)) = N

 

César Lozada