HYACINTHOS 28751

[Tran Quang Hung]:

 

 

Dear Mr Antreas,

 

{Qa,Qb,Qc} is Perspective to the 4th Euler triangle [Hyacinthos 28747] can be generalized as follows:

 

Let ABC be a triangle and P a point. 

 

IaIbIc is the excentral triangle.

 

AaAbAc is the pedal triangle of Ia.

 

The reflection of the line IaP in the sides of the triangle AaAbAc bound a triangle which has circumcenter Qa..

 

Define similarly the points Qb and Qc.

 

QaQbQc is an Euler triangle.

 

Then the triangle QaQbQc and the 4th Euler triangle are perspective. 

 

Which is the perspector in terms of P?

 

[Peter Moses]:

Hi Antreas,

b (a+b-c) c (a-b+c) (b+c)^2 p^2+c (a+b+c) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) p q-a (a-b-c) (a+b-c) c (-b^2+a c+c^2) q^2+b (a+b+c) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) p r+a (a+b+c) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) q r-a b (a-b-c) (a-b+c) (a b+b^2-c^2) r^2 : :

A point P{p,q,r} is on Feuerbach circumhyperbola of the tangential triangle the perspector will be on Euler line.

Examples:

------------------------------------------------------------------------
P = X(6) -> 

=  EULER LINE INTERCEPT OF X(12)X(3931)

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

= 5 X[1698] - X[2941]

 

= lies on the lines: {2, 3}, {12, 3931}, {115, 119}, {120, 5512}, {127, 25640}, {321, 17757}, {339, 21664}, {355, 1834}, {496, 5716}, {517, 1211}, {946, 3454}, {952, 17015}, {1213, 1766}, {1245, 21935}, {1329, 5955}, {1698, 2941}, {2345, 3820}, {3936, 5603}, {5016, 24390}, {5019, 5475}, {5706, 5810}, {5752, 5799}, {5886, 17056}, {5887, 10974}, {7682, 17052}, {7951, 17594}, {10175, 12618}, {10381, 24474}, {14672, 20621}

= complement of X(4221)

= midpoint of X(4) and X(4220)
= {X(429),X(442)}-harmonic conjugate of X(21530)
= X(i)-complementary conjugate of X(j) for these (i,j): {3420, 1125}, {9107, 8062}
= X(9058)-Ceva conjugate of X(523)

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

P = X(159) -> 

=  EULER LINE INTERCEPT OF X(12)X(18588)

(b + c)*(-a^2 + b^2 + c^2)*(a^6*b - a^4*b^3 - a^2*b^5 + b^7 + a^6*c + 2*a^5*b*c + 3*a^4*b^2*c + 2*a^3*b^3*c + a^2*b^4*c - b^6*c + 3*a^4*b*c^2 - 3*b^5*c^2 - a^4*c^3 + 2*a^3*b*c^3 + 3*b^4*c^3 + a^2*b*c^4 + 3*b^3*c^4 - a^2*c^5 - 3*b^2*c^5 - b*c^6 + c^7) : : 

 

= lies on these lines: {2, 3}, {12, 18588}, {115, 18591}, {119, 127}, {120, 14672}, {339, 1234}, {1060, 17720}, {3822, 18589}, {4463, 17757}, {7951, 10319}, {10202, 18635}

= complement of X(4227)
= X(i)-complementary conjugate of X(j) for these (i,j): {998, 942}, {9058, 8062}
------------------------------------------------------------------------

P = X(195) -> 

=  EULER LINE INTERCEPT OF X(12)X(3743)

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

 

= lies on these lines: {2, 3}, {12, 3743}, {80, 1834}, {119, 137}, {1089, 3704}, {1211, 3878}, {1213, 16548}, {3454, 11813}, {5443, 17056}, {20625, 25640}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {5, 1904, 11113}
= X(26711)-Ceva conjugate of X(523)
------------------------------------------------------------------------

P = X(399) -> 

=  EULER LINE INTERCEPT OF X(12)X(502)

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

 

= lies on the cubic K720 and these lines: {2, 3}, {12, 502}, {115, 3290}, {119, 3258}, {120, 5099}, {125, 517}, {515, 6739}, {523, 1577}, {758, 10693}, {1211, 10176}, {1213, 16547}, {1290, 5080}, {1737, 8287}, {3454, 25079}, {3814, 5520}, {4872, 23674}, {5074, 21253}, {8286, 30384}, {9956, 30436}, {16177, 25640}

= complement of X(1325)
= midpoint of X(1290) and X(5080)
= reflection of X(5520) and X(3814)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {2, 16430, 7483}, {5, 25648, 4187}, {5, 27687, 442}, {429, 21530, 442}, {1113, 1114, 2915}, {1312, 1313, 442}, {5142, 24933, 5}, {27553, 27685, 3}, {27554, 27686, 4}, {27555, 27687, 5}, {27561, 27693, 22}, {27562, 27694, 23}, {27578, 27715, 20}, {27579, 27716, 21}, {27580, 27717, 25}, {27581, 27718, 27}, {27582, 27721, 376}, {27583, 27722, 377}, {27584, 27723, 384}
= circumcircle inverse of X(2915)
= nine point circle inverse of X(442)
= polar circle inverse of X(28)
= orthoptic circle of the Steiner inellipe inverse of X(4220)
= complement of the isogonal of X(10693)
= X(i)-complementary conjugate of X(j) for these (i,j): {2766, 8062}, {10693, 10}
= X(i)-Ceva conjugate of X(j) for these (i,j): {1290, 523}, {5080, 758}
= crosssum of X(184) and X(19622)
= crossdifference of every pair of points on line {647, 1333}
------------------------------------------------------------------------

P = X(1740) -> 

=  EULER LINE INTERCEPT OF X(12)X(3743)

a^5*b^2 - a^3*b^4 + a^2*b^5 - b^7 + 2*a^5*b*c - a^3*b^3*c - a*b^5*c + a^5*c^2 + 2*a^3*b^2*c^2 + a^2*b^3*c^2 + 2*b^5*c^2 - a^3*b*c^3 + a^2*b^2*c^3 + 2*a*b^3*c^3 - b^4*c^3 - a^3*c^4 - b^3*c^4 + a^2*c^5 - a*b*c^5 + 2*b^2*c^5 - c^7  :: 

 

= lies on these lines: {2, 3}, {12, 3743}, {80, 1834}, {119, 137}, {1089, 3704}, {1211, 3878}, {1213, 16548}, {3454, 11813}, {5443, 17056}, {20625, 25640}

= {X(5),X(1904)}-harmonic conjugate of X(11113)
= X(26711)-Ceva conjugate of X(523)
------------------------------------------------------------------------

P = X(3216) -> 

=  EULER LINE INTERCEPT OF X(12)X(4424)

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

 

= lies on these lines: {2, 3}, {12, 4424}, {115, 22425}, {517, 3454}, {952, 1834}, {1211, 5690}, {1482, 3936}, {3073, 20575}, {3814, 24850}, {5510, 9955}, {5901, 17056}, {7680, 12621}, {7951, 24851}, {10974, 14988}, {12610, 21245}

= {X(5),X(5499)}-harmonic conjugate of X(15973)
= X(15617)-complementary conjugate of X(1125)
------------------------------------------------------------------------

Best regards,
Peter Moses.