Θέμα: [EMHL] Distance from centroid G and incenter I
- Dear friends,
after many efforts, I found a good proof for the
inequality IA+IB+IC <= GA+GB+GC (1)
based on an interesting property of X(177).
Let A'B'C' be the cevian triangle of the incenter I,
let A1B1C1 be the intouch triangle of ABC and
let A2B2C2 be the triangle with vertices on the
incircle and the segments IA, IB, IC that are the
mid-arc points of triangle A'B'C'.
The line A1A2 is a internal bisector of triangle
and is perpendicular to B2C2.
Hence A1A2, B1B2, C1C2 pass through the
incenter Q of A'B'C' that is the orthocenter of
This point Q is the point X(177) in ETC.
If c < b and M1 is the mid-point of BC then
BA1 = s-b < BA' = ac/(b+c) < BM1 = a/2.
Hence the point Q is on the other side of the centroid
relative to the line AA'.
The lines AA', BB', CC' devide the triangle ABC in 6
Tba = IBA', Tcb = ICB', Tac = IAC'
Tca = IA'C, Tab= IB'A, Tbc = IC'B
If the points G, I coincide then (1) is equality.
If G is on a bisector of ABC then Q is on the same
and the angle GIQ = pi.
If e.g. G is inside triangle Tbc then Q is inside
and since angle BIC is obtuse then also angle GIQ
is also obtuse and hence always
the dot product of vectors IG, IQ
is negative or zero i.e
IG.IQ <= 0 (2)
Let i, j, k be the unit vectors along IA, IB, IC
and r be the inradius of ABC.
The points Q, I are the orthocenter and circumcenter
and for vectors Sylvester's theorem gives
IA2 + IB2 + IC2 = IQ or
r(i + j + k) = IQ or
i + j + k = (1/r)IQ
IA + IB + IC = IA.i + IB.j + IC.k
= (IG +GA).i + (IG + GB).j + (IG + GC).k
= IG.(i + j + k) + GA.i + GB.j + GC.k
= (1/r)IG.IQ + GA.i + GB.j + GC.k
<= GA.i + GB.j + GC.k due to (2)
IA + IB + IC <= | GA.i + GB.j + GC.k |
<= | GA.i | + | GB.j | + | GC.k |
<= GA + GB + GC.
> > ABC with centroid G and incenter I, then___________________________________________________________
> > AG+BG+CG-AI-BI-CI=0 or >
> > 0. Is that true?
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