[Peter Moses]:
Hi Antreas,
An intouch version.
A’B’C’ = intouch triangle.
(Ja) = incircle of AB’C’
circle tangent externally to (Ja) (Jb) & (Jc), center
a ((b-c) (a^4 b-a^3 b^2-a^2 b^3+a b^4+a^4 c+a^3 b c+a^2 b^2 c-a b^3 c-2 b^4 c-a^3 c^2+a^2 b c^2+2 a b^2 c^2+2 b^3 c^2-a^2 c^3-a b c^3+2 b^2 c^3+a c^4-2 b c^4)+b (b-c) c (2 a^3+a^2 b-2 a b^2-b^3+a^2 c+2 a b c+5 b^2 c-2 a c^2+5 b c^2-c^3) Sin[A/2]-a (c (a^3 b+2 a^2 b^2-a b^3-2 b^4+a^3 c-a^2 b c+5 a b^2 c+3 b^3 c-a^2 c^2-5 a b c^2+b^2 c^2-a c^3-3 b c^3+c^4) Sin[B/2]- b (a^3 b-a^2 b^2-a b^3+b^4+a^3 c-a^2 b c-5 a b^2 c-3 b^3 c+2 a^2 c^2+5 a b c^2+b^2 c^2-a c^3+3 b c^3-2 c^4) Sin[C/2]))::
a ((b-c) (a^4 b-a^3 b^2-a^2 b^3+a b^4+a^4 c+a^3 b c+a^2 b^2 c-a b^3 c-2 b^4 c-a^3 c^2+a^2 b c^2+2 a b^2 c^2+2 b^3 c^2-a^2 c^3-a b c^3+2 b^2 c^3+a c^4-2 b c^4)+b (b-c) c (2 a^3+a^2 b-2 a b^2-b^3+a^2 c+2 a b c+5 b^2 c-2 a c^2+5 b c^2-c^3) Sin[A/2]-a (c (a^3 b+2 a^2 b^2-a b^3-2 b^4+a^3 c-a^2 b c+5 a b^2 c+3 b^3 c-a^2 c^2-5 a b c^2+b^2 c^2-a c^3-3 b c^3+c^4) Sin[B/2]- b (a^3 b-a^2 b^2-a b^3+b^4+a^3 c-a^2 b c-5 a b^2 c-3 b^3 c+2 a^2 c^2+5 a b c^2+b^2 c^2-a c^3+3 b c^3-2 c^4) Sin[C/2]))::
on lines {{164,5708},{177,942},{5049, 8422},{5439,11691},{8083,8091} ,{9957,11191}},
is tangent to the incircle of A’B’C’ at
Sin[A/2] ((a-b-c) (a^2 b-b^3+a^2 c+2 a b c+b^2 c+b c^2-c^3) Sin[A/2]+2 a b c ((a-b+c) Sin[B/2]+(a+b-c) Sin[C/2]))]::
on lines {{1,7597},{57,3659},{65,2089}, {174,354},{177,942}}.
is tangent to the incircle of A’B’C’ at
Sin[A/2] ((a-b-c) (a^2 b-b^3+a^2 c+2 a b c+b^2 c+b c^2-c^3) Sin[A/2]+2 a b c ((a-b+c) Sin[B/2]+(a+b-c) Sin[C/2]))]::
on lines {{1,7597},{57,3659},{65,2089}, {174,354},{177,942}}.
Best regards,
Peter Moses.
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου