#2575
Dear all
I would post this message several months ago!
I post it so late!
This message introduces a new geometry of two ordered triangles.
It contains a pair of points as orthocenters for two triangles.
Given two triangles ABC and DEF and an angle θ,
if P is a point and lines p_A,p_B,p_C are lines AP,BP,CP rotated by θ(counterclockwise),
suppose parallel lines to p_A,p_B,p_C through D,E,F respectively concur at another point P_θ.
The loci of P and P_θ are circumconics p and p_θ of triangles ABC and DEF respectively.
As P moves along p through A,B,C, P_θ moves along p_θ through A_θ, B_θ,C_θ.
As P_θ moves along p_θ through D,E,F, P moves along p through D_-θ,E_-θ,F_-θ.
Similarly, we get circumconics u and u_θ of D,E,F,A_-θ,B_-θ,C_-θ and A,B,C,D_θ,E_θ,F_θ respectively.
Circumonic p and u_θ of triangle ABC concur at the fourth point H_ABC.
Circumonic u and p_θ of triangle DEF concur at the fourth point H_DEF.
Points H_ABC and H_DEF remain fixed as θ varies.
Moreover, H_ABC and H_DEF behave like orthocenters of triangles.
That is, the fourth points of triangles H_ABC,B,C and H_DEF,E,F are A and D respectively.
the fourth points of triangles A,H_ABC,C and D,H_DEF,F are B and E respectively.
the fourth points of triangles A,B,H_ABC and D,E,H_DEF are C and F respectively.
Areas of three triangles ABC, D_θE_θF_θ, D_-θE_-θF_-θ are equal.
Areas of three triangles DEF, A_θB_θC_θ, A_-θB_-θC_-θ are equal.
If θ=0 or π/2, triangles A_θB_θC_θ and A_-θB_-θC_-θ coincide and D_θE_θF_θ and D_-θE_-θF_-θ do, too.
Suppose barycentrics of ABC and DEF are
A(1:0:0) B(0:1:0) C(0:0:1) D(dx:dy:dz) E(ex:ey:ez) F(fx:fy:fz)
a b c are "squares" of sidelengths BC,CA,AB.
First barcentric of H_ABC is (dx + dy + dz) / (-c (fx dy - dx fy) (dx ey - ex dy) + (a - b) (fx dy - dx fy) (dy ez - ey dz) + (-a + b + c) (fx dy - dx fy) (dz ex - ez dx) + c (fy dz - dy fz) (dx ey - ex dy) - a (fy dz - dy fz) (dy ez - ey dz) - (c - a) (fy dz - dy fz) (dz ex - ez dx) + b (dy ez - ey dz) (fz dx - dz fx) - b (dz ex - ez dx) (fz dx - dz fx))
To be continued...
Best regards,
Tsihong Lau
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#2589
Dear Tsihong Lau
A particular case: DEF=FaFbFc=Feuerbach triangle
H_ABC = X(5620)
H_DEF = (2r+R)(rR(2r+5R)-2(2r+R)s^2)) X(12) - (2R(2r^3+7rR(r+R)-(2r+R)s^2) X(5620)
H_DEF = ((b+c)(-a^14 (b+c) -a^13 (b^2+c^2) +a^12 (5 b^3+7 b^2 c+7 b c^2+5 c^3) +2 a^11 (2 b^4+3 b^3 c+6 b^2 c^2+3 b c^3+2 c^4) -a^10 (11 b^5+15 b^4 c+16 b^3 c^2+16 b^2 c^3+15 b c^4+11 c^5) -a^9 (5 b^6+19 b^5 c+35 b^4 c^2+30 b^3 c^3+35 b^2 c^4+19 b c^5+5 c^6) +a^8 (15 b^7+18 b^6 c+2 b^5 c^2-3 b^4 c^3-3 b^3 c^4+2 b^2 c^5+18 b c^6+15 c^7) +a^7 b c (24 b^6+43 b^5 c+26 b^4 c^2+22 b^3 c^3+26 b^2 c^4+43 b c^5+24 c^6) +a^6 (-15 b^9-19 b^8 c+14 b^7 c^2+26 b^6 c^3+13 b^5 c^4+13 b^4 c^5+26 b^3 c^6+14 b^2 c^7-19 b c^8-15 c^9) +a^5 (5 b^10-18 b^9 c-40 b^8 c^2-5 b^7 c^3+12 b^6 c^4+2 b^5 c^5+12 b^4 c^6-5 b^3 c^7-40 b^2 c^8-18 b c^9+5 c^10) +a^4 (11 b^11+15 b^10 c-19 b^9 c^2-38 b^8 c^3-10 b^7 c^4+21 b^6 c^5+21 b^5 c^6-10 b^4 c^7-38 b^3 c^8-19 b^2 c^9+15 b c^10+11 c^11) -a^3 (b^2-c^2)^2 (4 b^8-10 b^7 c-21 b^6 c^2-7 b^5 c^3+5 b^4 c^4-7 b^3 c^5-21 b^2 c^6-10 b c^7+4 c^8) -a^2 (b-c)^2 (b+c)^3 (5 b^8-8 b^6 c^2-7 b^5 c^3+16 b^4 c^4-7 b^3 c^5-8 b^2 c^6+5 c^8) +a (b^2-c^2)^4 (b^6-3 b^5 c-4 b^4 c^2+4 b^3 c^3-4 b^2 c^4-3 b c^5+c^6) +(b-c)^6 (b+c)^7 (b^2-b c+c^2)) : ... : ...),
with (6 - 9 - 13) - search numbers (0.181640280798688, -4.92623466068555, 6.96730065585960).
H_DEF lies on lines X(i)X(j) for these {i, j}: {12, 5620}, {101, 5949}, {442, 5953}.
Image and more information in:
http://amontes.webs.ull.es/otrashtm/HGT2017.htm#HG290817
Best regards,
Angel Montesdeoca
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#2600
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