HYACINTHOS 26501

[Le Viet An]:

Let ABC be a triangle, TaTbTc the pedal triangle of I (intouch triangle) and P a point different from I.

Denote:

Pa = the reflection of P in AI.
Ba, Ca = the orthogonal projections of of B, C on IPa.
La = the Euler line of TaBaCa.
Similarly Lb, Lc.

Then La, Lb, Lc are concurrent.

If P = O, H, G, Fe..... which is the point of concurrence?

 

 

[César Lozada]:

For P=u:v:w, the point of concurrence of the Euler lines is:

Q(P) = ((b^2-c^2)*(b-c)*a*u^2*(-a+b+ c)+(-(b-c)*(a^4-b*(2*b+c)*a^2+ b*(b+c)*(2*b-c)*a-(b^2-c^2)^2) *v+(b-c)*(a^4-c*(b+2*c)*a^2-c* (b+c)*(b-2*c)*a-(b^2-c^2)^2)* w)*u-(-b+c+a)*(c*a^3+b*(b+c)* a^2+(b^3+b^2*c-c^3)*a-(b^2-c^ 2)*(b-c)*b)*v^2+((b+c)*a^4+2*( b^2+c^2)*a^3-(2*b-c)*(b-2*c)*( b+c)*a^2-(2*b^2+3*b*c+2*c^2)*( b-c)^2*a+(b^2-c^2)*(b-c)^3)*w* v-(-c+a+b)*(b*a^3+c*(b+c)*a^2- (b^3-b*c^2-c^3)*a-(b^2-c^2)*( b-c)*c)*w^2)/a : :

 

Q(P) lies on the circle through ETC’s 11, 125, 1365. It has center X(11263) and squared radius

rad^2 = R*(2*r*(3*s^2-8*R^2-r^2)+(s^2- 11*r^2)*R)/(4*(2*r+3*R)^2)

 

All points on the line {I,P} have the same Q.

 

Q( P on the line IO )  = X(11)

Q( P on the line {1, 229} ) = X(125)

 

Particular points Q(P):

Q( O ) = X(11)

 

Q( H) = a*((b+c)*a^8-b*c*a^7-(b+c)*(3* b^2-b*c+3*c^2)*a^6+b*c*(b^2-b* c+c^2)*a^5+(b+c)*(3*b^4+3*c^4- b*c*(b^2-7*b*c+c^2))*a^4+b*c*( b^4-4*b^2*c^2+c^4)*a^3-(b+c)*( b^6+c^6+(b^4+c^4+3*b*c*(b-c)^ 2)*b*c)*a^2-(b^4+c^4-3*b*c*(b- c)^2)*(b+c)^2*b*c*a+(b^3+c^3)* (b^2-c^2)^2*b*c)*(b-c)^2 : : (barys)

= On lines: {124, 125}, {1364, 1365}

= [ 3.453089183640025, 3.11360440784455, -0.108641269818784 ]

 

Q( G ) = a*(b-c)^2*((b+c)*a^5+(b^2-b*c+ c^2)*a^4-(b^2-c^2)*(b-c)*a^3-( b^4+b^2*c^2+c^4)*a^2-2*b*c*(b+ c)*(b^2-4*b*c+c^2)*a+b*c*(b^2- 3*b*c+c^2)*(b+c)^2) : : (barys)

= On lines: {125, 2776}, {1086, 1357}, {2802, 3178}

= [ -0.388765975906126, 1.00880435432236, 3.121691917794805 ]

 

Q( K ) = (b-c)^2*(a^5+(b+c)*a^4-(b+2*c) *(2*b+c)*a^3+2*(b+c)^3*a^2-(3* b^4+3*c^4+b*c*(b-c)^2)*a+(b^4- c^4)*(b-c)) : : (barys)

= On lines: {125, 2775}, {512, 4904}, {1358, 1365}

= [ 0.205656524474334, 1.97825861025970, 2.176182432739509 ]

 

César Lozada