## Two Emile Hyacinthe Lemoine centers

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• Dear Hyacinthists looking at the beginning of the very interesting from EH Lemoyne : Suite de theoremes et de resultats concernant la geometrie du triangle
Message 1 of 3 , Dec 1, 2003
Dear Hyacinthists
looking at the beginning of the very interesting from EH Lemoyne :
Suite de theoremes et de resultats concernant la geometrie du
triangle
avalaible at
http://www.hti.umich.edu/u/umhistmath/
I see that he discovered the following centers :
1) Ma is the unique point on the segment BC such as
MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
cyclically. Then MaMbMc and ABC are perspective at the point of the
line IG with trilinear 1+2R/a:...
2) Na is the unique point on the segment BC such as
NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc
cyclically. Then NaNbNc and ABC are perspective at the point of the
line IG with trilinear (2R/a)-1:...
I don't think - but, may be, I'm wrong - that these points are in
the current ETC.
Friendly. Jean-Pierre
• Dear Jean-Pierre ... You mean Lemoine (as in the subject line) ... The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac. Define A = BAc / CAb.
Message 2 of 3 , Dec 2, 2003
Dear Jean-Pierre

>looking at the beginning of the very interesting from EH Lemoyne:

You mean Lemoine (as in the subject line)

>Suite de theoremes et de resultats concernant la geometrie du
>triangle avalaible at http://www.hti.umich.edu/u/umhistmath/
>I see that he discovered the following centers :
>1) Ma is the unique point on the segment BC such as
>MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
>cyclically. Then MaMbMc and ABC are perspective at the point of the
>line IG with trilinear 1+2R/a:...

The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac.

Define

A' = BAc /\ CAb. Similarly B', C'.

If my midnight calculations were correct, then these three
points are collinear.

And probably the same is true for the other case below
but didn't try it.

Good night from Athens

Antreas

>2) Na is the unique point on the segment BC such as
>NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc
>cyclically. Then NaNbNc and ABC are perspective at the point of the
>line IG with trilinear (2R/a)-1:...
>I don't think - but, may be, I'm wrong - that these points are in
>the current ETC.
>Friendly. Jean-Pierre

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• Dear Jean - Pierre ... In other words: Let Ma be *the* point on the line segment BC such that: MaB + MaMab = MaC + MaMac where Mab, Mac are the orthogonal
Message 3 of 3 , Dec 3, 2003
Dear Jean - Pierre

You wrote:
>I see that he [=Lemoine]discovered the following centers :
>1) Ma is the unique point on the segment BC such as
>MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
>cyclically. Then MaMbMc and ABC are perspective at the point of the
>line IG with trilinear 1+2R/a:...

In other words:

Let Ma be *the* point on the line segment BC such that:

MaB + MaMab = MaC + MaMac

where Mab, Mac are the orthogonal projections of Ma on
AB, AC, resp. Similarly Mb, Mc.

The triangles ABC, MaMbMc are perspective with trilinear perspector

((1 + sinA) / sinA ::)

Now,

1. Let Ka be *the* point on the line segment BC such that:

KaB + BKab = KaC + CKac

where Kab, Kac are the orthogonal projections of Ka on
AB, AC, resp. Similarly Kb, Kc.

The triangles ABC, KaKbKc are perspective with trilinear perspector:

((1 + cosA) / sinA ::) = (cot(A/2) ::) [mittenpunkt]

2. Let La be *the* point on the line segment BC such that:

LaLab + BLab = LaLaC + CLac

where Lab, Lac are the orthogonal projections of La on
AB, AC, resp. Similarly Lb, Lc.

The triangles ABC, LaLbLc are perspective with trilinear perspector:

((sinA + cosA) / sinA ::) = (1 + cotA ::)

3. Let Pa be *the* point on the line segment BC such that:

perimeter(BPaPab) = perimeter(CPaPac)

where Pab, Pac are the orthogonal projections of Pa on
AB, AC resp. Similarly Pb, Pc.

The triangles ABC, PaPbPc are perspective with trilinear perspector:

((1 + sinA + cosA) / sinA ::)

4. Let Qa be *the* point on the line segment BC such that:

area(BQaQab) = area(CQaQac)

where Qab, Qac are the orthogonal projections of Qa on
AB, AC, res. Similarly Qb, Qc.

The triangles ABC, QaQbQc are perspective with trilinear perspector:

(sqrt(cotA) ::)

>2) Na is the unique point on the segment BC such as
>NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc

That is, NaB - d(Na, AB) = NaC - d(Na,AC)

>cyclically. Then NaNbNc and ABC are perspective at the point of the
>line IG with trilinear (2R/a)-1:...

I think that we get similar results (for 1 and 2) just by changing
the "+" to "-" in the above trilinears.

Antreas
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