HYACINTHOS 25623

[Antreas P. Hatzipolaks]:

Let ABC be a triangle and P a point.

Denote:

A', B', C' = the reflections of P in BC, CA, AB, resp.

Which is the locus of P such that the NPCs of HPA', HPB', HPC' are coaxial ?

The Euler line?

If yes, which is the locus of the 2nd intersection (other than the midpoint of HP) as P moves on the Euler line?

 

 

[César Lozada]:

Locus = {Euler line} ∪ {circumcircle } ∪ { L∞ }

 

For P on Euler line, the 2^nd point of intersection Q(P) lies on the line {113, 403, 1986, 3580, 11557}. If OP/OH=p then

X(113)Q(P)/X(113)X(403)=q, where q=p*(OH/R)^2

 

ETC pairs (P,Q(P)): (3,1511), (4,1986), (5,11561), (23,3581), (186,186), (3520,3043), (7464,323)

 

Q( G ) = midpoint of X(110) and X(3060)

= (a^4-b^4+b^2*c^2-c^4)*((b^2+c^ 2)*a^4-2*(b^4-b^2*c^2+c^4)*a^ 2+(b^4-c^4)*(b^2-c^2))*a : : (trilinears)

= X(110)+2*X(1112), 2*X(113)+X(1986), X(113)+2*X(11557), 2*X(143)+X(5609), X(146)+2*X(974), X(399)+2*X(12236),   X(1539)+2*X(11561), X(1986)-4*X(11557), 3*X(5640)-X(9140), X(9143)+3*X(11002)

= on lines: {2,2781}, {23,6593}, {25,110}, {51,542}, {52,10294}, {74,9818}, {113,403}, {125,5133}, {143,5609}, {146,974}, {265,11818}, {381,5640}, {399,12236}, {511,5642}, {541,9730}, {568,5655}, {1495,11649}, {1511,2070}, {1539,11561}, {1550,11751}, {1992,2854}, {1995,9970}, {3448,7394}, {3796,10117}, {3917,5972}, {5095,8681}, {5422,5622}, {5621,10601}, {5643,12006}, {9517,9979}, {9729,10990}

= midpoint of X(i) and X(j) for these {i,j}: {110,3060}, {568,5655}, {5890,10706}

= reflection of X(i) in X(j) for these (i,j): (125,5943), (3060,1112), (3917,5972), (9140,12099)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (113,11557,1986), (5640,9140,12099)

= [ -0.740277326486018, -1.48298604941201, 5.009013589878544 ]

 

For P on the circumcircle, the locus of P (three-petals-flower, magenta curve in attached sketch)  is the quartic with trilinear parametric equation F(a,b,c,t) : F(b,c,a,t) : F(c,a,b,t)

Where

F(a,b,c,t)= (a-b)*(-c*t+a+b+c)*(a-c)*(-b* t+a+b+c)*(a^2*((b^2+c^2)*a^4- 2*b*c*(b+c)*a^3-2*(b^4+c^4-b* c*(b^2+b*c+c^2))*a^2+2*(b^2-c^ 2)*(b-c)*b*c*a+(b^2-c^2)^2*(b- c)^2)*t^2-2*a*(a+b+c)*((b+c)* a^5-4*b*c*a^4-(b+c)*(b^2-3*b* c+c^2)*a^3-(b^2+c^2)*(b-c)^2* a^2+(b^2-c^2)*(b-c)*b*c*a+(b^ 2-c^2)^2*(b-c)^2)*t+(2*a^6-2*( b+c)*a^5+(b^2+c^2)*a^4-2*(b^2- c^2)*(b-c)*a^3+2*b*c*(b-c)^2* a^2+(b^2-c^2)^2*(b-c)^2)*(a+b+ c)^2)*b*c

 

Q(X(98)) =

= ((b^4+c^4)*a^4-2*(c^6+b^6)*a^ 2+b^8+4*b^4*c^4+c^8-2*b^2*c^6- 2*b^6*c^2)*a/((b^2+c^2)*a^2-b^ 4-c^4) : : (trilinears)

= On cubic K289 and  lines: {4,512}, {23,1976}, {98,385}, {232,1692}, {895,11653}, {5640,5967}

= [ -8.256732964779022, -0.68192956401297, 7.923569394583679 ]

 

Q(X(99)) =

= a*((b^4+c^4)*a^4-2*(c^6+b^6)* a^2+b^8+4*b^4*c^4+c^8-2*b^2*c^ 6-2*b^6*c^2)/(b^2-c^2) : : (trilinears)

= On lines: {4,69}, {99,512}, {112,249}, {526,9182}, {924,4590}, {2715,4611}, {2855,9160}, {9181,10411}

= [ 2.803938486474784, -0.07773306549389, 2.400354225799479 ]

 

César Lozada