HYACINTHOS 27192

[Antreas P. Hatzipolakis]:

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

Denote:

A", B", C" = the reflections of A', B', C' in L, resp.

A*,B*, C* = the reflections of A, B, C in PA", PB", PC", resp.

P lies on the Euler line of A*B*C*.

Which is the point P wrt triangle A*B*C*? 


[César Lozada]:

 

Suppose OP/OH=t and let O*, H* be the circumcenter, orthocenter of A*B*C*. Then 

O*P/O*H* = R^2/((9*R^2-2*SW)*t)

 

 

[.....]


[APH]:

Thanks,  César  !

A question is for which point(s) P the Euler lines of ABC and A*B*C* are perpendicular (at P) ?

 
[César Lozada]:


In a given triangle, the direction of the Euler line of A*B*C* does not change.  

The angle θ between both Euler lines is:

θ  = arctan( -(SB-SC)*(SC-SA)*(SA-SB)/(2*S* (S^2+18*(3*R^2-SW)*R^2+SW^2)) )

 
[APH]:

..... which means the Euler lines of the triangles A*B*C*, as P moves on the Euler line, are parallels, concurrent at a fixed point on the Line at infinity.

Is it, and its isogonal conjugate on the circumcircle, interesting points?

[César Lozada]:

 

>..... which means the Euler lines of the triangles A*B*C*, as P moves on the Euler line, are parallels, concurrent at a

> fixed point on the Line at infinity.

The point in infinity is:

 

Zi = Infinity point of line X(3)X(476)

= S^4+(-3*R^2*(9*R^2+3*SA-4*SW)+ 2*SA^2+SB*SC-SW^2)*S^2+(27*R^ 2*(3*R^2-SW)+SW^2)*SB*SC : : (barys)

= (10*cos(2*A)+cos(4*A)+9)*cos( B-C)-4*cos(A)*cos(2*(B-C))+1/ 2*cos(3*(B-C))-12*cos(A)-4* cos(3*A) : : (trilinears)

= on lines: {3, 476}, {4, 14670}, {5, 3258}, {30, 511}, {381, 15111}, {382, 14264}, {399, 14480}, {1511, 7471}, {3447, 12028}, {5609, 14611}, {6070, 10264}, {9179, 14650}, {10095, 12052}, {10620, 14508}, {14851, 14993}, {14895, 15807}

= isogonal conjugate of Zo

= [ 1.2167671022145610, 1..0462219533702310, -1.2858923225861110 ]

 

whose isogonal conjugate is:

Zo = X(476)X(5663) ∩ X(477)X(526)

= (SB+SC)*(S^4+(-3*R^2*(9*R^2+3* SB-4*SW)+2*SB^2+SC*SA-SW^2)*S^ 2+(27*R^2*(3*R^2-SW)+SW^2)*SC* SA)*(S^4+(-3*R^2*(9*R^2+3*SC- 4*SW)+2*SC^2+SA*SB-SW^2)*S^2+( 27*R^2*(3*R^2-SW)+SW^2)*SA*SB) : : (barys)

= ((10*cos(2*B)+cos(4*B)+9)*cos( -C+A)-4*cos(B)*cos(-2*C+2*A)+ 1/2*cos(-3*C+3*A)-12*cos(B)-4* cos(3*B))*((10*cos(2*C)+cos(4* C)+9)*cos(A-B)-4*cos(C)*cos(2* A-2*B)+1/2*cos(3*A-3*B)-12* cos(C)-4*cos(3*C)) : : (barys)

= on the circumcircle and on these lines: {74, 14809}, {107, 7722}, {110, 14933}, {476, 5663}, {477, 526}

= isogonal conjugate of Zi

= antipode of Z’o in the circumcircle

= [ 11.3608807356090300, 13.2128234231196900, -10.7501582274565000 ]

 

Zo has antipode:

Z’o = X(476)X(526) ∩ X(477)X(5663)

= sin(A-B)*sin(A-C)*((cos(A-B)- 2*cos(C))^2-(cos(2*C)+1/2)^2)* ((cos(A-C)-2*cos(B))^2-(cos(2* B)+1/2)^2) : : (trilinears)

= on the circumcircle and on these lines: {476, 526}, {477, 5663}, {1294, 12219}

= isogonal conjugate of Z’i

= antipode of Zo in the circumcircle

= [ 2.2038450527836560, -1.4229215696568820, 3.6086047750005330 ]

 

and Z’o has isogonal conjugate:

Z’i = Infinity point of line X(476)X(10412)

= sin(B-C)*((cos(B-C)-2*cos(A))^ 2-(cos(2*A)+1/2)^2) : :  (trilinears)

= on lines: {30, 511}, {351, 9158}, {476, 10412}, {3134, 3258}, {14380, 14989}

= isogonal conjugate of Z’o

= [ 1.9917590179867390, -3.0848701374214040, 1.2164059329901700 ]

 

César Lozada