ADGEOM 4214 * ADGEOM 4219

#4214

Dear geometers,

 

Let ABC be a triangle.

 

A'B'C' is Morley triangle.

 

A1, A2 divide AB', AC' in the ratio k.

 

B1, B2 divide BC', BA' in the ratio k.

 

C1, C2 divide CA', CB' in the ratio k.

 

Then Euler lines of the triangle AA1A2, BB1B2, CC1C2 are concurrent at a point P.

 

And P lies on a fixed line when k changes, which is this line ?

 

Best regards,

Tran Quang Hung.

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#4219

 

 

[César Lozada]:

This concurrence occurs for several triangles in the index of triangles referenced in ETC and the point of concurrence and either Q is a fixed point or Q always lies on a fixed line.

The appearance of (T, K) in the following list means that for triangle T, Q is always X(K):

(anti-Ara, 4), (anti-Ascella, 3), (1st anti-Brocard, 98), (1st anti-circumperp, 3), (anti-McCay, 671), (circumorthic, 3),  (hexyl, 1), (Hutson intouch, 1), (incircle-circles, 1), (intouch, 1), (Johnson, 4), (medial, 30), (6th mixtilinear, 1),

 

The appearance of (T, {I, J}, K) in the following list means that for triangle T Q lies on the line {X(i), X(j)}, whose trilinear pole is X(K):

(ABC-X3 reflections, {2, 3}, 648), (1st circumperp, {3, 8}, 13136), (outer-Garcia, {8, 79}, P1), (1st Morley, {357, 1136}, P2), (2nd Morley, {1134, 1136}, P3), (3rd Morley, {357, 1134}, P4), (orthic, {4, 74}, P5)

 

where:

P1 = trilinear pole of the line X(8)X(79)

= b^2*c^2*(a-b)*(a-c)*(a^2+c*a-b^2+c^2)* (a^2+b*a+b^2-c^2): : (trilinears)

= on lines: {2, 94}, {476, 9070}, {645, 4585}, {662, 1577}, {3699, 6742}, {8287, 14616}

= isotomic conjugate of X(14838)

= trilinear pole of the line {8, 79}

= [ 0.0771664861892042, -0.0321132886105630, 3.6272814573197530 ]

 

P2 = trilinear pole of the line X(357)X(1136)

= (2*cos(2*B/3-2*Pi/3)-1)*(2*cos(2*C/3-2*Pi/3)-1)*csc((B-C)/3) : : (trilinears)

= on lines {}

= trilinear pole of the line X(357)X(1136)

= [ -0.2124839000637748, 0.2333219507863389, 3.5772033929309630 ]

 

P3 = trilinear pole of the line X(1134)X(1136)

= (2*cos(2*B/3)-1)*(2*cos(2*C/3)-1)*csc((B-C)/3) : : (trilinears)

= on lines { }

= trilinear pole of the line X(1134)X(1136)

= [ -0.4129180249109538, 0.4073395077229288, 3.5492377572889450 ]

 

P4 = trilinear pole of the line X(357)X(1134)

= (2*cos(2*B/3-4*Pi/3)-1)*(2*cos(2*C/3-4*Pi/3)-1)*csc((B-C)/3) : : (trilinears)

= on lines {}

= trilinear pole of the line {357, 1134}

= [ 0.7568322443459275, -0.6082414680076237, 3.7124475392915360 ]

 

P5 = trilinear pole of the line X(4)X(74)

= SB^2*SC^2*(SA-SB)*(SA-SC) *(S^2-3*SA*SB)*(S^2-3*SA*SC) : : (barycentrics)

= on lines: {107, 523}, {525, 648}, {685, 879}, {1494, 6330}, {1503, 10152}, {1897, 4064}, {1990, 14165}, {2394, 2404}, {3267, 6331}, {3470, 13450}, {6530, 9139}, {14618, 15352}

= isogonal conjugate of X(1636)

= polar conjugate of X(9033)

= trilinear pole of the line {4, 74}

= [ -0.5805934377880054, 0.5098612495022659, 3.5556498189234290 ]

 

César Lozada

2017-11-29