- 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

>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

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

>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:...

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

- Dear Jean - Pierre

You wrote:>I see that he [=Lemoine]discovered the following centers :

In other words:

>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:...

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

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

>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

I think that we get similar results (for 1 and 2) just by changing

>line IG with trilinear (2R/a)-1:...

the "+" to "-" in the above trilinears.

Antreas

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