HYACINTHOS 28771

[Tran Quang Hung]:

 

Let ABC be a triangle inscribed in circle (O).

 

H is orthocenter.

 

A', B', C' are the reflections of H through BC, CA, AB, resp.

 

Let (wbc) be the circle tangent to segment HB, HC' and (O) internally.

 

Let (wcb) be the circle tangent to segment HC, HB' and (O) internally.

 

Let A* be the outer homothety center of (wbc) and (wcb).

 

Define similarly the points B* and C*.

 

Then A*, B* and C* are collinear on line d.

 

Tripole of d lies on Euler line of ABC. Which is this point?

 

 

[Peter Moses]:

 

Hi Antreas,

>Let (wbc) be the circle tangent to segment HB, HC' and (O) internally.  

center = {a (b-c) (a^2-b^2-c^2)+(a+b-c) (a-b+c) S,(-a^2+b^2+c^2) (a b+S),(a^2-b^2-c^2) (a c-S)};
radius = ((a^2-b^2-c^2) (a (a+b-c) (a-b+c)-2 (b-c) S))/(2 (a-b-c) (a+b+c) S).
midpoint wbc & wcb = A-vertex of the Fuhrmann triangle

radical trace (wbc) & (wcb) = {a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4,b (b+c) (-a^2+b^2+c^2),c (b+c) (-a^2+b^2+c^2)}.
the 3 radical traces are perspective with the intouch triangle at:

=  X(1)X(1361)∩X(7)X(286)

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

 

= lies on the cubic K714 and on these lines: {1, 1361}, {7, 286}, {12, 26932}, {34, 26892}, {49, 23070}, {51, 1393}, {56, 58}, {57, 1745}, {63, 7066}, {65, 515}, {73, 22345}, {84, 7355}, {109, 23850}, {181, 1454}, {185, 7004}, {201, 3917}, {221, 22654}, {226, 14058}, {227, 8679}, {388, 26871}, {603, 1425}, {942, 1875}, {971, 1887}, {1038, 3784}, {1122, 1439}, {1214, 11573}, {1354, 1363}, {1355, 1367}, {1358, 20618}, {1400, 14597}, {1406, 18954}, {1433, 3304}, {1455, 23842}, {1469, 7289}, {1473, 19349}, {2003, 19365}, {3220, 26888}, {4014, 7702}, {4303, 22341}, {5399, 23981}, {5907, 24430}, {7352, 24467}, {9291, 18026}, {15524, 20323}, {17114, 18838}, {18915, 26929}, {26933, 26955}

 

= X(18180)-Ceva conjugate of X(1393)
= X(i)-isoconjugate of X(j) for these (i,j): {8, 2190}, {9, 275}, {33, 95}, {41, 276}, {54, 318}, {78, 8884}, {212, 8795}, {281, 2167}, {312, 8882}, {933, 4086}, {2148, 7017}, {2289, 8794}, {4041, 18831}, {8611, 16813}
= crosspoint of X(7) and X(222)
= crosssum of X(55) and X(281)
= barycentric product X(i) X(j) for these {i,j}: {5, 222}, {7, 216}, {51, 348}, {53, 1804}, {56, 343}, {63, 1393}, {65, 16697}, {73, 17167}, {77, 1953}, {217, 6063}, {278, 5562}, {324, 7335}, {331, 418}, {603, 14213}, {604, 18695}, {1214, 18180}, {1397, 28706}, {1625, 17094}, {1813, 21102}, {2179, 7182}, {2181, 7183}, {3199, 7055}, {4565, 6368}, {4573, 15451}, {7069, 7177}, {7178, 23181}, {8798, 18623}, {17076, 27372}
= barycentric quotient X(i) / X(j) for these {i,j}: {5, 7017}, {7, 276}, {51, 281}, {56, 275}, {216, 8}, {217, 55}, {222, 95}, {278, 8795}, {343, 3596}, {418, 219}, {603, 2167}, {604, 2190}, {608, 8884}, {1118, 8794}, {1393, 92}, {1397, 8882}, {1953, 318}, {2179, 33}, {3199, 1857}, {4565, 18831}, {5562, 345}, {7069, 7101}, {7335, 97}, {15451, 3700}, {16697, 314}, {18695, 28659}, {23181, 645}
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): {56, 222, 7335}, {1425, 3937, 603}

>Let A* be the outer homothety center of (wbc) and (wcb).
= {-(b-c) (-a^2+b^2-c^2) (a^2+b^2-c^2),(b+c) (a^2+b^2-c^2) (-a^2+b^2+c^2),(b+c) (-a^2+b^2-c^2) (-a^2+b^2+c^2)}.

>Tripole of d lies on Euler line of ABC. Which is this point?

X(27).

Best regards,
Peter Moses.