HYACINTHOS 27095

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and A'B'C' the antipedal triangle of P.

Denote:

Aa, Ab, Ac = the orthogonal projections of A' on AO, BO, CO, resp.

Ba, Bb, Bc = the orthogonal projections of B' on AO, BO, CO, resp.

Ca, Cb, Cc = the orthogonal projections of C' on AO, BO, CO, resp.

La, Lb, Lc = the Euler lines of AaAbAc, BaBbBc, CaCbCc, resp.

1. P = H  [A'B'C' = the antimedial triangle]
La, Lb, Lc concur on the circumcircle.
The parallels to La, Lb, Lc though A, B, C, resp. are concurrent.

2. P = O [ A'B'C' = the tangential triangle]
[Ab = B, Ac = C etc The triangles are AaBC, BbCA, CcAB]
A*B*C*, ABC are parallelogic.
The parallelogic center (ABC, A*B*C*) lies on the circumcircle

 

3. P = I [A'B'C' = the excentral triangle]
A*B*C*, ABC are parallelogic.

 

 

[César Lozada]:

 

I assume A*B*C* is the triangle bounded by La, Lb, Lc.

 

> 1. P = H  [A'B'C' = the antimedial triangle]
> La, Lb, Lc concur on the circumcircle.   
> The parallels to La, Lb, Lc though A, B, C, resp. are concurrent. 

 

At X(930) and X(513), resp.

 

> 2. P = O [ A'B'C' = the tangential triangle]
> [Ab = B, Ac = C etc The triangles are AaBC, BbCA, CcAB]
> A*B*C*, ABC are parallelogic.
> The parallelogic center (ABC, A*B*C*) lies on the circumcircle

 

ABC->A*B*C = X(1141)

A*B*C*->ABC = trilinear pole of the line X(577)X(1147)

= SA*(SA-SB)*(SA-SC)*(S^2+SA* SB)*(S^2+SA*SC)*(SB+SC)^2 : :  (barys)

= on lines: {54, 5504}, {97, 3917}, {110, 933}, {252, 5449}, {930, 6368}, {1157, 13754}, {2071, 3484}

= trilinear pole of the line {577, 1147}

= barycentric product X(i)*X(j) for these {i, j}: {54, 4558}, {69, 14586}, {97, 110}, {99, 14533}, {394, 933}, {525, 14587}, {662, 2169}, {2148, 4592}, {2167, 4575}, {10411, 11077}

= barycentric quotient X(i)/X(j) for these (i, j): (48, 2618), (54, 14618), (69, 15415), (97, 850), (110, 324), (112, 13450), (184, 12077), (577, 6368), (933, 2052), (1576, 53), (2169, 1577), (2623, 2970), (4558, 311), (4575, 14213), (11077, 10412), (14533, 523), (14573, 2489), (14574, 3199), (14585, 15451), (14586, 4), (14587, 648)

= trilinear product X(i)*X(j) for these {i, j}: {54, 4575}, {63, 14586}, {97, 163}, {110, 2169}, {255, 933}, {656, 14587}, {662, 14533}, {2148, 4558}

= trilinear quotient X(i)/X(j) for these (i, j): (3, 2618), (48, 12077), (97, 1577), (162, 13450), (163, 53), (255, 6368), (304, 15415), (662, 324), (933, 158), (1576, 2181), (2148, 2501), (2167, 14618), (2169, 523), (2616, 2970), (4558, 14213), (4575, 5), (4592, 311), (14533, 661), (14586, 19), (14587, 162)

= [ 3.1454484338236380, 4.7158029214260270, -1.0758675869983960 ]

 

> 3. P = I [A'B'C' = the excentral triangle]
> A*B*C*, ABC are parallelogic.

This is not true with my assumption.

 

César Lozada