HYACINTHOS 29137

 

[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

 

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[Ercole Suppa]:

 

 

Hi Antreas,

 

Let Q = Q(d1) be the concurrency point of the lines DX, EY, FZ. 

 

We have:

 

*** if d1 = HO (Euler line) then 

 

Q(HO) = X(3258)

 

*** if d1 = HI then 

 

Q(HI) = X(10017)

 

*** if d1 = HK then 

 

Q(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) : : (barys)

 

= 3*X[2]-X[2867] 

 

= 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)

 

= complementary conjugate of X(2881)

 

= reflection of X(1562) in X(14117)

 

= (6-9-13)  search numbers:  [2.4512998320928429880, 3.3877200492212084435, 0.1639506791729995708]

 

 

 

Best regards,

Ercole Suppa