HYACINTHOS 28963

[Vu Thanh Tung]:

 

Dear César Lozada and other friends,

Let K_AK_BK_C be the pedal triangle of a point  K w.r.t triangle ABC.

K_{AB}, K_{AC} are the projection of K_A on CA, AB resp.

Define K_{BC}, K_{BA}, K_{CA}, K_{CB} similarly.

A'B'C' is the triangle bounded by the lines K_{BC}K_{CB}, K_{CA}K_{CA}, K_{AB}K_{BA}.

 

We can prove that AK is perpendicular to K_{BC}K_{CB}. The line  K_{BC}K_{CB} is the line B'C' so AK is perpendicular to B'C'

Similarly BK is perpendicular to C'A' and CK is perpendicular to A'B'.

Thus the triangle ABC is orthologic to triangle A'B'C' with one orthology center K. 

The other orthogoy K' is the intersection of 3 lines through A',B',C'  perpendicular to BC, CA, AB.

The triangle A'B'C' is similar to the antipedal of K w.r.t ABC.

 

If  K=X(13) of ABC then angle BKC =  angle CKA = angle AKB = 120°  => angle B'A'C' =  angle A'C'B' = angle C'B'A' =   60°

  => the triangle A'B'C' is equilateral.

Similarly for K=X(14).

Best regards,

Vu Thanh Tung

[César Lozada]:

 

Dear Mr Thanh,

 

Apologizes. I read “cevian” instead of “pedal”.

 

Triangles T13 (using K=X(13)) and T14 (using K=X(14)) are indeed equilateral. Their centers are:

 

·         For T13:

T13 has squared-side: 1/9*S^2*(2*S^2+sqrt(3)*(3*R^2+SW)*S+SW^2)^2/R^4/(SW+sqrt(3)*S)^3

 

Center:

O’ = 6*(9*R^2-3*SA+SW)*S^2-sqrt(3)*(4*S^2+3*(9*SA-11*SW)*R^2+3*SA^2-6*SB*SC-SW^2)*S-3*(SB+SC)*(3*R^2*(3*SA-2*SW)-SA^2+SB*SC) : :  (barys)

= lies on these lines: {395, 1506}, {530, 15544}, {8838, 9112}

= [ 0.8368244322180914, 1.2620823133352760, 2.3806885270362230 ]

 

T13 is homothetic to these triangles: 3rd Fermat-Dao, 7th Fermat-Dao, 11th Fermat-Dao, 15th Fermat-Dao, 1st isodynamic-Dao, 3rd isodynamic-Dao, 1st Lemoine-Dao, outer-Napoleon

 

·         For T14:

T14 has squared-side: 1/9*S^2*(2*S^2-sqrt(3)*(3*R^2+SW)*S+SW^2)^2/R^4/(SW-sqrt(3)*S)^3

 

Center:

O” = 6*(9*R^2-3*SA+SW)*S^2+sqrt(3)*(4*S^2+3*(9*SA-11*SW)*R^2+3*SA^2-6*SB*SC-SW^2)*S-3*(SB+SC)*(3*R^2*(3*SA-2*SW)-SA^2+SB*SC) : :  (barys)

= lies on these lines: {396, 1506}, {531, 15544}, {8466, 16397}, {8836, 9113}

= [ -0.5742916682506035, -0.4725102408682428, 4.2328446493934430 ]

 

T14 is homothetic to these triangles: 4th Fermat-Dao, 8th Fermat-Dao, 12th Fermat-Dao, 16th Fermat-Dao, 2nd isodynamic-Dao, 4th isodynamic-Dao, 2nd Lemoine-Dao, inner-Napoleon

 

T13 and T14 are orthologic with many traingles, including ABC. Also, they are parallelogic with many other triangles. Unfortunately, coordinates of orthologic and parallelogic centers are complicated.

 

César Lozada