HYACINTHOS 26655

 

[APH]:

 

In the first Interbalkan Math. Conference (Athens 1934), the Greek mathematician Demetrios Ghiokas announced a proof of Morley's theorem (he doesn't mention Morley's name!). He proves it as a corollary of the general theorem:

Let ABC be a triangle and let AK, AL be isogonal lines from A ie BAK = LAC = a [in the figure is a typo: is d instead of a <actually there is no typo, simply a looks like d>] and similarly BK', BL' isogonal lines from B ie CBK' = ABL' = b and CK", CL" isogonal lines from C ie ACK' = BCL" = c [I added K', L'.K".L" in the figure]

Denote:

AK /\ BL' = P, BK' /\ CL" = Q, CK"/\ AL = R and

 

AL /\ BK = P', BL' /\ CK" = Q', CL" /\ AK = R'

The triangles PQR, P'Q'R' are perspective.

 

Reference:

Sur un theoreme de la theorie du triangle par. M. D. Ghiocas (Athenes)
Actes Congres Interbalkan. Math., Athenes 1934, p. 103-104

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Rewriting the theorem:

Let ABC be a triangle.

Let AAb, AAc be two isogonal cevians from A, ie angles BAAb = CAAc = α [: lower case Greek letter alpha]

 

BBc, BBa two isogonal  cevians from B, ie angles CBBc = ABBa = β [: lower case Greek letter beta]

CCa, CCb rwo isogonal cevians from C, ie angles ACCa = BCCb = γ [: lower case Greek letter gamma].

Denote:

BBc /\ CCb = A'

CCa /\ AAc = B'

AAb /\ BBa = C'

BBa /\ CCa = A"

CCb /\ AAb = B"

AAc /\ BBc = C"

The triangles A'B'C', A"B"C" are perspective.

Loci (complicated !)

1. Let α = β = γ : = ω [ : lower case Greek letter omega]

Which is the locus of the perspector of A'B'C', A"B"C" as ω varies ?

2. Let α = A/t, β = B/t, γ = C/t

Which is the locus of the perspector of A'B'C', A"B"C" as t varies ?

 

Antreas P. Hatzipolakis

 

 

 

[César Lozada]:

 

Please let me replace ω with θ

 

1)      A’B’C’ and ABC are perspective with perspector

Q’ = sin(A)/sin(A-α) : :   (trilinears)

 

2)      A”B”C” and ABC are perspective with perspector

Q” = sin(A-α)/sin(A) : :   (trilinears) = isogonal conjugate of Q’

 

3)      A’B’C’ and A”B”C” are perspective with perspector

Q = sin(α)*sin(B-β)*sin(C-γ) + sin(A-α)*sin(β)*sin(γ) : : (trilinears)

 

I would expect that α, β, γ should be symmetrical quantities for such perspectivities, ie, α=f(a,b,c), β=f(b,c,a) and γ=f(c,a,b) for some homogeneous function f,  but surprisingly (at least for me), these three perspectivities occur for any arbitrary values of  α, β, γ.  Moreover, Q, Q’ and Q” are collinear.

 

Particular case:

If α = β = γ = θ ≠ 0 then

a)       Q(θ) = sin(θ)*sin(A- θ)+sin(B- θ)*sin(C- θ) : :  (trilinears)

= cos(2*θ)*cos(A)+cos(B)*cos(C) : : (trilinears)

b)      Q(θ) lies on the Euler line of ABC and Q(θ)=2*cos(2* θ)*X(3)+X(4) 

c)       Q(-θ) = Q(θ) and Q(θ+k*Pi)=Q(θ), for any integer k

d)      Q(Pi/2-θ) = Q(k*Pi/2+θ)

e)      Q(Pi/2-θ) and Q(Pi/2+θ) are harmonic conjugates w/r to O and H

f)        Q(Pi/3) = X(30) && Q(Pi)=X(2)

g)       If P(t) is on the Euler line of ABC such that OP/OH=t, the required θ for Q(θ)=P(t) satisfies cos(θ)^2=(t+1)/(4*t). Therefore, the actual range of the transformation θ ->P is {Euler line}-segment (X(20), X(3)), X(20) excluded.  

 

Some Q(θ):

 

Q(Pi/12) = sqrt(3)*cos(A)+2*cos(B)*cos(C) : : (trilinears)

= sqrt(3)*X(3)+X(4)  = X(4)-3*X(2044)

= On lines: {2, 3}, {15, 3071}, {16, 3070}, {17, 615}, {18, 590}, {371, 398}, {372, 397}, {395, 8960}, {623, 642}, {624, 641}, {1151, 5339}, {1152, 5340}, {1587, 11486}, {1588, 11485}, {5318, 6396}, {5321, 6200}, {5334, 6221}, {5335, 6398}, {5343, 6449}, {5344, 6450}, {5365, 6455}, {5366, 6456}, {8976, 11489}, {8981, 11543}, {11488, 13951}, {11542, 13966}

= midpoint of X(2041) and X(3529)

= reflection of Q(5*Pi/12) in X(550)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2, 2041, 632), (3, 4, Q(5*Pi/12)), (3, 382, 2043), (4, 2046, 5), (376, 5073, Q(5*Pi/12)), (382, 3522, Q(5*Pi/12)), (472, 1593, 4185), (1884, 1904, 9714), (2567, 6861, 24), (3530, 3858, Q(5*Pi/12)), (3843, 10299, Q(5*Pi/12)), (5576, 7495, Q(5*Pi/12)), (7524, 13737, 7412)

= [ 2.222058865220567, 1.34667709493024, 1.682783939930878 ]

 

Q(Pi/6) = X(5)

 

Q(Pi/4) = X(4)

 

Q(Pi/3) = X(30)

 

Q(5*Pi/12) = -sqrt(3)*cos(A)+2*cos(B)*cos( C) : : (trilinears)

= -sqrt(3)*X(3)+X(4) = X(4)-3*X(2043)

= On lines: {2, 3}, {15, 3070}, {16, 3071}, {17, 590}, {18, 615}, {371, 397}, {372, 398}, {396, 8960}, {623, 641}, {624, 642}, {1151, 5340}, {1152, 5339}, {1587, 11485}, {1588, 11486}, {5318, 6200}, {5321, 6396}, {5334, 6398}, {5335, 6221}, {5343, 6450}, {5344, 6449}, {5365, 6456}, {5366, 6455}, {8976, 11488}, {8981, 11542}, {11489, 13951}, {11543, 13966}

= midpoint of X(2042) and X(3529)

= reflection of Q(Pi/12) in X(550)

= Inverse of X(7734) in the radical circle of Stammler circles

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (3, 4, Q(Pi/12)), (3, 382, 2044), (4, 2045, 5), (376, 5073, Q(Pi/12)), (381, 3523, Q(Pi/12)), (382, 3522, Q(Pi/12)), (2041, 2043, 3), (2041, 2045, 4), (3134, 13737, 6983), (3135, 3546, 6913), (3530, 3858, Q(Pi/12)), (3843, 10299, Q(Pi/12)), (5576, 7495, Q(Pi/12)) = [ 23.801649228295010, 22.86933995374934, -23.177332052978250 ]

 

Q(Pi/2) = X(20)

 

 

 

2. Let α = A*t, β = B*t, γ = C*t  (let’s avoid divisions by zero)

> Which is the locus of the perspector of A'B'C', A"B"C" as t varies ?

 

Perspectivities and expressions of perspectors are kept.  Therefore Morley theorem is proved.

Locus?: I don’t know which is the locus of Q, but I can assure that it is not a straight line.

 

César Lozada