HYACINTHOS 25437

[Tran Quang Hung]:

 

Let ABC be a triangle with incenter I.

A line d cuts IA,IB,IC at A',B',C'.

 P is reflection of I through d.

 Then the circles (PAA'),(PBB'),(PCC') are coaxial and the second intersection lies on the circumcircle (O) of ABC.

 

[Antreas P. Hatzipolakis]:

Dear Tran,

Extraversions:

Let ABC be a triangle and IaIbIc the excentral triangle.

A line d intersects the internal bisector of A [ = AIa] at Aa, the external bisector of B [ = BIa] at Ab and the external bisector of C [ = CIa] at Ac.

Let Pa be the reflection of Ia in d.

The circumcircles of PaAAa,PaBAb,PaCAc are coaxial.

Let A* be the the second intersection. Similarly B*, C* for Ib, Ic. resp.

[César Lozada]:

 

 

Let d = [L, M, N] (trilinears). Then the 2^nd point of intersection A*(d) is:

A*(d) = f(a, b, c, A, B, C, L, M, N)  :  g(a, b, c, A, B, C, L, M, N)  :  g(a, c, b, A, C, B, L, N, M) 

where

f(a, b, c, A, B, C, L, M, N) = a/((-2*a+b+c)*L^2+2*L*a*(cos(C )*M+cos(B)*N)-(b+c)*(M^2-2*M*N *cos(A)+N^2))

g(a, b, c, A, B, C, L, M, N) = b/((-a+2*b+c)*M^2-2*M*b*(cos(C )*L+cos(A)*N)+(a-c)*(L^2-2*cos (B)*L*N+N^2))

 

and cyclic expressions for B*(d) and C*(d)

 

A*(d), B*(d) and C*(d) lie on the circumcircle of ABC.

----------------------------

If d is the trilinear polar of Q=u:v:w (trilinears) then:

A*(Q) = F(a, b, c, A, B, C, u, v, w)  :  G(a, b, c, A, B, C, u, v, w)  :  G(a, c, b, A, C, B, u, w, v) 

where

F(a, b, c, A, B, C, u, v, w) = a/((-2*a+b+c)*v^2*w^2+2*a*u*v* w*(cos(B)*v+cos(C)*w)-(b+c)*(v ^2-2*cos(A)*v*w+w^2)*u^2)

G(a, b, c, A, B, C, u, v, w) = b/((-a+2*b+c)*u^2*w^2-2*b*(cos (A)*u+cos(C)*w)*u*w*v+(a-c)*( u^2-2*cos(B)*u*w+w^2)*v^2)

 

and cyclic expressions for B*(Q) and C*(Q).

----------------------------

For any Q, triangle A*B*C* is perspective to anti-Mandart-incircle and 1^st circumperp triangles, with perspectors Zm(Q) and Zc(Q), respectively (general forms given below). No perspectors were found in ETC.

 

No triangles being always orthologic o parallelogic to A*B*C* were found. In each case, 190 defined triangles were checked.

 

A pair of perspectors:

 

Zm( I ) =  {55,2316}/\{3196,6600}

a*(-a+b+c)*(2*a^5-4*(b+c)*a^4+ (2*b^2+9*b*c+2*c^2)*a^3+(b+c)* (2*b^2-9*b*c+2*c^2)*a^2-(4*b^4 +4*c^4-b*c*(6*b^2-b*c+6*c^2))* a+2*(b^3+c^3)*(b-c)^2) : :  (trilinears)

= On lines: {55,2316}, {3196,6600}

= [ -5.875141187750339, 13.12298826821923, -2.732877617128782 ]

 

Zc( I ) = midpoint of X(165) and X(9355)

3*a^4-4*(b+c)*a^3-(b^2-9*b*c+c ^2)*a^2+(b+c)*(4*b^2-9*b*c+4*c ^2)*a-2*(b^2-c^2)^2 : : (trilinears)

= 2*(4*R^2-15*R*r-4*r^2)*X(9)-(8 *R^2+6*R*r+r^2-3*s^2)*X(48)

= On lines: {3,3196}, {6,10247}, {9,48}, {44,517}, {45,10246}, {165,2246}, {952,4370}, {1635,2827}, {1743,2170}, {1766,3973}, {2291,2348}, {2792,10175}

= midpoint of X(165) and X(9355)

= [ 11.586988692305010, 14.20925108265187, -11.544350279453840 ]

------------------------------

The locus of Q such that A*B*C* and ABC are perspective is  :

{CyclicSum[a*v*w*(b*c*v*w-(-a^ 2+b^2+c^2)*u^2)]=0} \/

{ CyclicSum[ (b-c)*a*v*w*((a-c)*(a-b)*b*c*v ^3*w^3+2*b*c*u^2*(-a^2+b^2+c^2 )*(v^2*w^2-u^2*(v^2+w^2))+(a^4 -2*(b^2+b*c+c^2)*a^2+2*(b+c)*b *c*a+c^4+b^4+4*b^2*c^2)*u^4*v* w) ]=0}

 

No ETC centers, other than G, were found on this locus. (The trilinear polar of G is the line-at-infinity)

------------------------------

General trilinear coordinates of perspectors:

Zm(Q) = (-a+b+c)*(a^2*(a^4-2*(b^2+b*c+ c^2)*a^2+2*b*c^3+c^4+b^4+2*b^3 *c)*u^4*w^2*v^2-a*(a+b-c)*(a^4 -a^3*c-b*(2*b-c)*a^2-(b^2-c^2) *c*a+(b+c)*(b^3-c^3))*u^3*v^3* w^2+(-a^6+2*(b+c)*a^5-(b+c)^2* a^4-(b+c)*(b^2+c^2)*a^3+2*(b^ 4+c^4)*a^2-(b^2-c^2)^2*(b+c)* a+2*(b^2-c^2)^2*b*c)*u^2*v^3*w ^3-a*(a-b+c)*(a^4-a^3*b+c*(b-2 *c)*a^2+(b^2-c^2)*b*a-(b+c)*(b ^3-c^3))*v^2*u^3*w^3+a^2*b*c*( a^2+(b-c)*a-b*(2*b-3*c))*u^4*w ^4+a^2*b*c*(a^2-(b-c)*a+c*(3*b -2*c))*u^4*v^4-a^2*b*c*(a^2-(b +c)*a+3*b*c)*v^4*w^4+a^2*((b-c )*a+c^2-2*b*c-b^2)*(-a^2+b^2+c ^2)*u^4*v^3*w-a^2*((b-c)*a+2*b *c-b^2+c^2)*(-a^2+b^2+c^2)*u^4 *v*w^3-a*b*(a^2-(b-2*c)*a+3*c* (b-c))*(a^2-b^2+c^2)*w*v^4*u^ 3+b*(a-c)*(a^4-2*(b^2-2*c^2)* a^2+(b^2-c^2)^2)*w^2*v^4*u^2+c *a*(a^2-(2*b+c)*a+b*(b+3*c))*( a^2+b^2-c^2)*w^4*v^3*u-c*a*(a^ 2+(2*b-c)*a-3*b*(b-c))*(a^2+b^ 2-c^2)*w^4*u^3*v+c*(a-b)*(a^4+ 2*(2*b^2-c^2)*a^2+(b^2-c^2)^2) *w^4*v^2*u^2+a*b*(a^2-(b+2*c)* a+c*(3*b+c))*(a^2-b^2+c^2)*w^ 3*v^4*u) : :

 

Zc(Q) = (a-c)*(a^2-c*a+b*c-2*c^2-b^2)* a^2*b*c*u^4*v^4-(-a^2+b^2+c^2) *((b-c)*a^2-2*b*c*a+3*b^2*c+b* c^2+c^3-b^3)*a^2*u^4*v^3*w+(-a +b+c)*(a^4-2*(b^2-b*c+c^2)*a^2 +b^4-2*b*c^3+c^4-2*b^3*c+8*b^2 *c^2)*a^2*u^4*v^2*w^2+(-a^2+b^ 2+c^2)*((b-c)*a^2+2*b*c*a-b^2* c-3*b*c^2+c^3-b^3)*a^2*u^4*v* w^3+(a-b)*(a^2-b*a+b*c-c^2-2* b^2)*a^2*b*c*u^4*w^4-(a-c)*(a^ 2-b^2+c^2)*(a^2-2*c*a+2*b*c-3* c^2-b^2)*a*b*u^3*v^4*w+(a^6+c* a^5-(3*b^2+b*c+c^2)*a^4+2*c^2* (b-c)*a^3+(3*b^4-c^4)*a^2-(b^ 2-c^2)*(b-c)^2*c*a-(b^2-c^2)*( b^4+c^4-b*c*(b^2-4*b*c-c^2)))* a*u^3*v^3*w^2+(a^6+b*a^5-(b^2+ b*c+3*c^2)*a^4-2*b^2*(b-c)*a^ 3-(b^4-3*c^4)*a^2+(b^2-c^2)*( b-c)^2*b*a+(b^2-c^2)*(b^4+c^4+ b*c*(b^2+4*b*c-c^2)))*a*u^3*v^ 2*w^3-(a-b)*(a^2-2*b*a+2*b*c- c^2-3*b^2)*(a^2+b^2-c^2)*a*c* u^3*v*w^4-(a-c)*(a-b+c)*(a^4- 2*(b^2-2*c^2)*a^2+(b^2-c^2)^2) *b*u^2*v^4*w^2+(a^4+(b+c)*a^3- 2*b*c*a^2-(b^2-c^2)*(b-c)*a-( b^2-c^2)^2)*(-a^2+b^2+c^2)*a*u ^2*v^3*w^3-(a-b)*(a+b-c)*(a^4+ 2*(2*b^2-c^2)*a^2+(b^2-c^2)^2) *c*u^2*v^2*w^4+(a-c)*(a^2+2*c* a-2*b*c+c^2-b^2)*(a^2-b^2+c^2) *a*b*u*v^4*w^3+(a-b)*(a^2+b^2- c^2)*(a^2+2*b*a-2*b*c-c^2+b^2) *a*c*u*v^3*w^4-(a-c)*(a-b)*(a+ b+c)*a^2*b*c*v^4*w^4 : :

 

César Lozada