HYACINTHOS 29138

 

[Corneliu Manescu-Avram]:

 

Let ABC be a triangle and d1, d2 two perpendicular lines intersecting each other at H.

 

Denote:

D, E, F = the orthogonal projections of A, B, C on d1. resp.
X, Y, Z = the orthogonal projections of A, B, C on d2 resp.

 

Then DX, EY, FZ are concurrent.

By Corneliu Manescu-Avram

 

Note:

As the line d1 moves around H, the point of concurrence moves on the NPC.
Like a line d when is moving around O: its orthopole moves on the NPC.

Question:
Which are the points of concurrence for some special d1's: Euler line, HI, HK, and possibly others passing through H.

APH

 

[Peter Moses]:

 

Hi Antreas,

Suppose a point, P{p, q, r}, is at infinity, then the two lines, d1 and d2, are line[H, P] and line[H, orthopoint(P)]
and the concurrence is
p (c^2 q^2 - b^2 r^2) ((a^2 - b^2 + c^2) q - (a^2 + b^2 - c^2) r) : :
on the NP circle and the cevian circle of P.

Examples:
HI -> X(10017).
HG -> X(3258).

HK -> COMPLEMENT OF X(2867)

 

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

 

= 3 X[6793] - 2 X[15639]  

= lies on the nine-point circle and these lines: {2, 2867}, {4, 32687}, {113, 15341}, {122, 647}, {125, 6587}, {127, 525}, {132, 1503}, {133, 5523}, {1560, 32125}, {2697, 2715}, {6794, 18809}, {13509, 18402}, {14401, 16177}, {17434, 20625}

= complement of X(2867)
= reflection of X(1562) in X(14117)
= reflection of X(1562) in the van Aubel line
= polar circle inverse of X(32687)
= Moses Radical circle inverse of X(122)
= complement of the isogonal of X(2881)
= orthic isogonal conjugate of X(2881)
= X(i)-complementary conjugate of X(j) for these (i,j): {1, 2881}, {2881, 10}
= X(4)-Ceva conjugate of X(2881)
= crosspoint of X(525) and X(1503)
= crosssum of X(112) and X(1297)
= crossdifference of every pair of points on line {2435, 2445}

HNa -> X(3259).
HX(9) -> X(1566).
HX(69) -> X(2679).

Best regards,
Peter Moses.