#1530
Dear friends,
Given a triangle ABC, its 1st Morley triangle DEF, and its Den Roussel triangle GHI, DEF and GHI are perspective at a point P.
The 1st trilinear of P
(6 Cos[B/3 - (5 C)/3] - 21 Cos[B/3 - C] - 12 Cos[B - C] + 48 Cos[B/3 - C/3] - 21 Cos[B - C/3] + 6 Cos[(5 B)/3 - C/3] - 57 Cos[B/3 + C/3] - 42 Cos[B + C/3] + 33 Cos[(5 B)/3 + C/3] - 3 Cos[(7 B)/3 + C/3] - 42 Cos[B/3 + C] + 96 Cos[B + C] + 12 Cos[(5 B)/3 + C] - 21 Cos[(7 B)/3 + C] + 6 Cos[3 B + C] + 33 Cos[B/3 + (5 C)/3] + 12 Cos[B + (5 C)/3] - 75 Cos[(5 B)/3 + (5 C)/3] + 30 Cos[(7 B)/3 + (5 C)/3] + 15 Cos[3 B + (5 C)/3] - 3 Cos[(11 B)/3 + (5 C)/3] - 3 Cos[B/3 + (7C)/3] - 21 Cos[B + (7 C)/3] + 30 Cos[(5 B)/3 + (7 C)/3] + 54 Cos[(7 B)/3 + (7 C)/3] - 36 Cos[3 B + (7 C)/3] - 6 Cos[(11 B)/3 + (7 C)/3] + 6 Cos[B + 3 C] + 15 Cos[(5 B)/3 + 3 C] - 36 Cos[(7 B)/3 + 3 C] - 15 Cos[3 B + 3 C] + 12 Cos[(11 B)/3 + 3 C] - 3 Cos[(5 B)/3 + (11 C)/3] - 6 Cos[(7 B)/3 + (11 C)/3] + 12 Cos[3 B + (11 C)/3] - 3 Cos[(11 B)/3 + (11 C)/3] - 2 Sqrt[3] Sin[B/3 - (5 C)/3] - 4 Sqrt[3] Sin[B - (5 C)/3] + 11 Sqrt[3] Sin[B/3 - C] + 4 Sqrt[3] Sin[(5 B)/3 - C] - 11 Sqrt[3] Sin[B - C/3] + 2 Sqrt[3] Sin[(5 B)/3 - C/3] - 47 Sqrt[3] Sin[B/3 + C/3] + 32 Sqrt[3] Sin[B + C/3] + 3 Sqrt[3] Sin[(5 B)/3 + C/3] + Sqrt[3] Sin[(7 B)/3 + C/3] + 2 Sqrt[3] Sin[3 B + C/3] + 32 Sqrt[3] Sin[B/3 + C] + 18 Sqrt[3] Sin[B + C] - 38 Sqrt[3] Sin[(5 B)/3 + C] + 5 Sqrt[3] Sin[(7 B)/3 + C] - 2 Sqrt[3] Sin[(11 B)/3 + C] + 3 Sqrt[3] Sin[B/3 + (5 C)/3] - 38 Sqrt[3] Sin[B + (5 C)/3] + 11 Sqrt[3] Sin[(5 B)/3 + (5 C)/3] + 30 Sqrt[3] Sin[(7 B)/3 + (5 C)/3] - 11 Sqrt[3] Sin[3 B + (5 C)/3] - Sqrt[3] Sin[(11 B)/3 + (5 C)/3] + Sqrt[3] Sin[B/3 + (7 C)/3] + 5 Sqrt[3] Sin[B + (7 C)/3] + 30 Sqrt[3] Sin[(5 B)/3 + (7 C)/3] - 28 Sqrt[3] Sin[(7 B)/3 + (7 C)/3] - 14 Sqrt[3] Sin[3 B + (7 C)/3] + 6 Sqrt[3] Sin[(11 B)/3 + (7 C)/3] + 2 Sqrt[3] Sin[B/3 + 3 C] - 11 Sqrt[3] Sin[(5 B)/3 + 3 C] - 14 Sqrt[3] Sin[(7 B)/3 + 3 C] + 21 Sqrt[3] Sin[3 B + 3 C] + 2 Sqrt[3] Sin[(11 B)/3 + 3 C] - 2 Sqrt[3] Sin[B + (11 C)/3] - Sqrt[3] Sin[(5 B)/3 + (11 C)/3] + 6 Sqrt[3] Sin[(7 B)/3 + (11 C)/3] + 2 Sqrt[3] Sin[3 B + (11 C)/3] - 5 Sqrt[3] Sin[(11 B)/3 + (11 C)/3])/(1+2Cos[2A/3])^2.
For triangle {6,9,13}, kx=1.19153490968316..
Best regards,
Seiichi Kirikami
P. S. There are 2 cases of computation.
(1) The concurrency determinant=0, even without A+B+C=Pi. Example, message #1526.
(2) The concurrency determinant=0, with A+B+C=Pi. Example, this message.
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#1545
Dear friends,
By chance, I found the simple coordinates for the homothetic center of the 1st Morley triangle and Roussel triangle in Hyacinthos message #8563.
Trilinears= = {(Cos[(Pi+2A)/3]+Cos[(Pi+2B)/3]+Cos[(Pi+2C)/3])Sin[(Pi+A)/3]+2(Sin[2B/3]Sin[2C/3]-Sin[A/3]Sin[(Pi-A)/3])Sin[A/3] : : }.
If anyone know the simple coordinates for the perspector of the 1st adjunct Morley triangle and Roussel triangle, I hope that they will be added to this message as a reply.
Best regards,
Seiichi Kirikami
P. S. Here I use Roussel triangle instead of Den Roussel triangle because it is used in Hyacinthos message #8563.
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#1546
Dear Seiichi,
I already had some expression for this perspector, but it was about twice as long as yours.
Triggered by your new expression I had another try:
-2 Sqrt[3] Sin[A/3] Sin[2 A/3] + Cos[A + Pi/6] + 2 (Cos[C/3] Cos[B + Pi/6] + Cos[B/3] Cos[C + Pi/6]) : :
Yet I am not satisfied with it.
I don't think it is the ultimate expression.
Maybe you or someone else can brew another simpler version.
Best regards,
Chris van Tienhoven
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#1569
Dear Chris,
Thank you very much for your reply.
(1) The center of Roussel triangle (Hyacinthos message #21406):
Trilinears={(Cos[B/3] (2 Cos[C/3] Sin[A] + Sin[B]) + Cos[C/3] Sin[C] + Cos[A/3] (Sin[A] + 2 Cos[C/3] Sin[B] + 2 Cos[B/3] Sin[C])) (2 Sin[B/3] Sin[C/3] - Sin[A/3 - Pi/6]) + 2 (Cos[A/3] + 2 Cos[B/3] Cos[C/3]) (2 Sin[A/3] (Sin[B] Sin[C/3] + Sin[B/3] Sin[C]) + Sin[A] (2 Sin[B/3] Sin[C/3] - Sin[A/3 - Pi/6]) - Sin[B] Sin[B/3 - Pi/6] - Sin[C] Sin[C/3 - Pi/6] ): : }. Trilinears=2 X(356)+X(3277).
(2) The homothetic center of the 1st Morley triangle and Roussel triangle (Hyacinthos message #8563 and ADGEOM message #1546 ):
Trilinears= = {(Cos[(Pi+2A)/3]+Cos[(Pi+2B)/3]+Cos[(Pi+2C)/3])Sin[(Pi+A)/3]+2(Sin[2B/3]Sin[2C/3]-Sin[A/3]Sin[(Pi-A)/3])Sin[A/3] : : } by Milorad Stevanovic. Trilinears= {-2Sqrt[3]Sin[A/3]Sin[2A/3]+Cos[A+Pi/6]+2(Cos[C/3]Cos[B+Pi/6]+Cos[B/3]Cos[C+Pi/6]) : : } by Chris van Tienhoven.
(3) The perspector of the 1st adjunct Morley triangle and Roussel triangle: Trilinears= (to be added).
Best regards, Seiichi Kirikami
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#1578
Dear Seiichi and Chris,
Another triangle to consider is the 'Adjunct Roussel' triangle, constructed analagously to the 1st Adjunct Morley triangle. It is perspective to the 1st Morley triangle at a non-ETC point on lines 15,358 3280,3602 (search 1.499634729695783), and to the Roussel triangle at a non-ETC point (search 2.427937089459433).
Best regards,
Randy Hutson
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#1580
And another related triangle:
Let La be the line through the circumcircle intercepts (other than A) of the internal angle trisectors of A, and define Lb, Lc cyclically. Let A' = Lb/\La, B' = Lc/\La, C' = La/\Lb.
A'B'C' is homothetic to ABC at a non-ETC point with trilinears cos A - cos(A/3) : :, on lines 3,358 357,3279 3280,3602 (at least), and search value 0.195460788980394.
A'B'C' is perspective to the Roussel triangle at the same point (search 2.427937089459433) as the perspector of the Roussel triangle and adjunct Roussel triangle.
Best regards,
Randy Hutson
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#1647
Dear friends,
Additions concerning points of Roussel triangle
[1] The center of Roussel triangle (Hyacinthos message #21406):
(1)Trilinears={(Cos[B/3] (2 Cos[C/3] Sin[A] + Sin[B]) + Cos[C/3] Sin[C] + Cos[A/3] (Sin[A] + 2 Cos[C/3] Sin[B] + 2 Cos[B/3] Sin[C])) (2 Sin[B/3] Sin[C/3] - Sin[A/3 - Pi/6]) + 2 (Cos[A/3] + 2 Cos[B/3] Cos[C/3]) (2 Sin[A/3] (Sin[B] Sin[C/3] + Sin[B/3] Sin[C]) + Sin[A] (2 Sin[B/3] Sin[C/3] - Sin[A/3 - Pi/6]) - Sin[B] Sin[B/3 - Pi/6] - Sin[C] Sin[C/3 - Pi/6] ): : }.
(2)Trilinears=2 X(356)+X(3277).
(3)Trilinears={-3 Sqrt[3] Cos[B/3 - C] - 2 Sqrt[3] Cos[B - C] + 2 Sqrt[3] Cos[B/3 - C/3] - 3 Sqrt[3] Cos[B - C/3] + 2 Sqrt[3] Cos[B/3 + C/3] + 2 Sqrt[3] Cos[B + C] + 3 Sqrt[3] Cos[(5 B)/3 + C] + 3 Sqrt[3] Cos[B + (5 C)/3] - 2 Cos[(5 B)/3 - C/3 - Pi/6] - 2 Cos[B/3 - (5 C)/3 + Pi/6] +Sin[B/3 - C] - Sin[B - C/3] + 4 Sin[B/3 + C/3] + Sin[(5 B)/3 + C] + Sin[B + (5 C)/3] +
6 Sin[(5 B)/3 + (5 C)/3] - 2 Sqrt[3] Sin[(5 B)/3 + C/3 + Pi/6] - 2 Sqrt[3] Sin[B/3 + (5 C)/3 + Pi/6]: : }.
[2] The homothetic center of the 1st Morley triangle and Roussel triangle (Hyacinthos message #8563 and ADGEOM message #1546 ):
(1)Trilinears= = {(Cos[(Pi+2A)/3]+Cos[(Pi+2B)/3]+Cos[(Pi+2C)/3])Sin[(Pi+A)/3]+2(Sin[2B/3]Sin[2C/3]-Sin[A/3]Sin[(Pi-A)/3])Sin[A/3] : : } by Milorad Stevanovic.
(2)Trilinears=
{-2Sqrt[3]Sin[A/3]Sin[2A/3]+Cos[A+Pi/6]+2(Cos[C/3]Cos[B+Pi/6]+Cos[B/3]Cos[C+Pi/6]) : : } by Chris van Tienhoven.
(3)Trilinears={4 Sqrt[3] Cos[B - C] + 8 Sqrt[3] Cos[B/3 + C/3] - 6 Sqrt[3] Cos[B + C/3] - 6 Sqrt[3] Cos[B/3 + C] + 5 Sqrt[3] Cos[B + C] + 5 Sqrt[3] Cos[(5 B)/3 + (5 C)/3] - 12 Cos[(5 B)/3 + C + Pi/6] - 12 Cos[B + (5 C)/3 + Pi/6] - 12 Sin[B + C/3] - 12 Sin[B/3 + C] - 3 Sin[B + C] - 3 Sin[(5 B)/3 + (5 C)/3] - 2 Sqrt[3] Sin[(5 B)/3 - C/3 - Pi/6] + 2 Sqrt[3] Sin[B/3 - (5 C)/3 + Pi/6]: : }.
[3] The perspector of the 1st adjunct Morley triangle and Roussel triangle:
(1)Trilinears={2 Cos[B - C] + 4 Cos[B/3 - C/3] - 2 Cos[(5 B)/3 + C/3] - 5 Cos[B + C] - 2 Cos[B/3 + (5 C)/3] + 2 Sqrt[3] Cos[B - C/3 - Pi/6] + 2 Sqrt[3] Cos[B/3 - C + Pi/6] + 2 Sqrt[3] Cos[B/3 + C/3 + Pi/6] - 2 Sqrt[3] Cos[(5 B)/3 + C + Pi/6] - 2 Sqrt[3] Cos[B + (5 C)/3 + Pi/6] + Sqrt[3] Sin[B + C]: : }.
The added expressions([1](3), [2](3), [3](1)) are linear combinations of sine and cosine functions.
Best regards,
Seiichi Kirikami
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#1658
Dear friends,
A shorter expression of the center of Roussel triangle:
Trilinears={- Cos[B/3 - C] - Cos[B - C/3] - Cos[B + C/3] - Cos[B/3 + C] + 3 Cos[B + C] + 2 Sin[B/3 + C/3 + Pi/6]): : }.
Best regards,
Seiichi Kirikami
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#1660
Dear Seiichi:
This is shorter:
-cos(A/3) + ( cos(B/3)*cos(C)+cos(C/3)*cos(B) ) + (3/2)*cos(A)
Regards
César Lozada
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#1663
Dear Cesar,
Thank you very much for your reduction.
I define the length of expression as expanded, a kind of sine-cosine series.
( For expansion, I use // TrigExpand//Trigfactor commands.)
The length of ours = 6.
Best regards,
Seiichi Kirikami
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#1728
Dear friends, dear Randy,
Let ABC be a triangle with its circumcircle k.
The trisector of angle A lying near AC intersects k at Ac.
The trisector of angle A lying near AB intersects k at Ab.
The trisector of angle B lying near BA intersects k at Ba.
The trisector of angle B lying near BC intersects k at Bc.
The trisector of angle C lying near CB intersects k at Cb.
The trisector of angle C lying near CA intersects k at Ca.
Denote the intersection of BBc and CCb by D, the intersection of CCa and AAc by E, and the intersection of AAb and BBa by F. DEF is the 1st Morley triangle.
Denote the intersection of AcCa and AbBa by D1, the intersection of BaAb and BcCb by E1, and the intersection of CbBc and CaAc by F1. D1E1F1 is the Roussel triangle.
Denote the intersection of AcBc and AbCb by D2, the intersection of BaCa and BcAc by E2, and the intersection of CbAb and CaBa by F2. D2E2F2 is the adjunct Roussel triangle.
The 1st Morley triangle DEF and the adjunct Roussel triangle D2E2F2 are perspective and have the following concurrent point.
Trilinears={-2Cos[5B/3+5C/3] + 2Sqrt[3]Cos[B+C+Pi/6] + 2Sin[B/3-C-Pi/6] + 2
Sin[B/3+C/3-Pi/6] - Sin[5B/3+C-Pi/6] - 2Sin[B+5C/3-Pi/6] – 2 Sin[B-C/3+Pi/6]: : }.
Its length=7.
Best regards,
Seiichi Kirikami
P. S. My computation showed that Roussel triangle D1E1F1 and adjunct Roussel triangle D2E2F2 were not perspective.
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#1730
Dear Seiichi,
Your construction of the adjunct Roussel triangle agrees with mine, but I do find that it is perspective to the Roussel triangle, and the perspector has ETC search value 2.427937089459433. It is also perspective to the 1st Morley triangle at ETC search=1.499634729695783 (on lines 15,358 and 3280,3602).
I am not sure what you mean by length=7.
Best regards,
Randy Hutson
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#1733
Dear Seiichi and friends,
Another triangle can be constructed from this configuration:
Denote the intersection of BcBa and CaCb by D3, the intersection of CaCb and AbAc by E3 and the intersection of AbAc and BcBa by F3. D3E3F3 is homothetic to ABC at trilinears cos A - cos(A/3) : : (Search=0.195460788980394, on lines 3,358 357,3279 3280,3602), and perspective to the Roussel triangle D1E1F1 and the adjunct Roussel triangle D2E2F2 at (search=2.427937089459433).
PS. Disregard my earlier question about length=7. I see you are referring to the number of terms in the coordinates.
Best regards,
Randy Hutson
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#1738
Dear Randy,
I was wrong. The Roussel and adjunct Roussel triangles are perspective.
I am trying to deduce the coordinates. But a kind of asymmetry in them prevents me from reaching the final solution now.
Best regards,
Seiichi Kieikami
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#1744
Dear Randy,
D3E3F3 is a remarkable triangle.
Especially because it is homothetic with ABC.
Let us generalize this construction:
Let D1.E1.F1 and D2.E2.F2 be two perspective triangles with perspector P.
* Ld = Line (D1E1^D2F2) ^ (D1F1^D2E2)
* Le = Line (E1F1^ E2D2) ^ (E1D1^E2F2)
* Lf = Line (F1D1^F2E2) ^ (F1E1^F2D2)
New triangle Ld.Le.Lf is perspective with triangles D1.E1.F1 as well as D2.E2.F2.
Their mutual common perspector is P.
As a consequence the perspector of D3E3F3 with Roussel triangle D1E1F1 and adjunct Roussel triangle D2E2F2 is the same as the perspector of D1E1F1 and D2E2F2.
Best regards,
Chris van Tienhoven
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#1762
Dear Randy,
The following is the coordinates of the perspector of Roussel and adjunct Roussel triangles.
Trilinears =
{(5Sqrt[3] + 4 Sqrt[3]Cos[2B/3] + 2 Sqrt[3]Cos[4B/3] + 4Sqrt[3]Cos[2B/3 - 2C/3] -
2Sqrt[3]Cos[2B/3 + 2C/3] - 5Sqrt[3]Cos[4B/3 + 2C/3] - Sqrt[3]Cos[2B + 2C/3] +
4Sqrt[3]Cos[2C/3] + 2 Sqrt[3]Cos[4C/3] - 5Sqrt[3]Cos[2B/3 + 4C/3] –
3Sqrt[3]Cos[4B/3 + 4C/3] - Sqrt[3]Cos[2B + 4C/3] -
Sqrt[3]Cos[2B/3 + 2C] - Sqrt[3]Cos[4B/3 + 2C] + Sqrt[3]Cos[2B + 2C] - 2Cos[8B/3 + 2C/3 + Pi/6] - 2Cos[2B/3 + 8C/3 + Pi/6] + 4Sin[4B/3] + 2Sin[2B] - 2Sin[2B/3 - 2C] + 2Sin[2B - 2C/3] + 7Sin[4B/3 + 2C/3] + 5Sin[2B + 2C/3] + 4Sin[4C/3] + 2Sin[2C] + 7Sin[2B/3 + 4C/3] + 11Sin[4B/3 + 4C/3] + 5Sin[2B + 4C/3] - 2Sin[8B/3 + 4C/3] + 5Sin[2B/3 + 2C] + 5Sin[4B/3 + 2C] - 5Sin[2B + 2C] - 2Sin[4B/3 + 8C/3] -
2Sqrt[3]Sin[2B/3 - 4C/3 - Pi/6] + 2Sqrt[3]Sin[4B/3 - 2C/3 + Pi/6] -
2Sqrt[3]Sin[8B/3 + 2C + Pi/6] - 2Sqrt[3]Sin[2B + 8C/3 + Pi/6] -
2Sqrt[3]Sin[8B/3 + 8C/3 + Pi/6])/Sin[A/3]: : }.
Its length=38.
There might be a shorter expression.
Best regards,
Seiichi Kirikami
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