## two Lemuan's points

Expand Messages
• Dear colleagues! Does anybody know the non projective proof of next fact. Let A B C is the circumcevian triangle of ABC with respect to point P, L, L are the
Message 1 of 9 , Jun 4, 2004
Dear colleagues!
Does anybody know the non projective proof of next fact.
Let A'B'C' is the circumcevian triangle of ABC with respect to point P, L,
L' are the Lemuan's point of ABC and A'B'C'. Then P is in LL'.
In fact if we concern projective transformations saving the circumcircle
then the Lemuan point is invariant. So it is sufficient to concern an
evident case when P=O.

Sincerely Alexey
• Dear Alexey ... point P, L, ... The harmonic homology with center P and axis the polar line of P wrt the circumcircle fixes the circumcircle and maps A,B,C,L
Message 2 of 9 , Jun 4, 2004
Dear Alexey
> Does anybody know the non projective proof of next fact.
> Let A'B'C' is the circumcevian triangle of ABC with respect to
point P, L,
> L' are the Lemuan's point of ABC and A'B'C'. Then P is in LL'.
The harmonic homology with center P and axis the polar line of P wrt
the circumcircle fixes the circumcircle and maps A,B,C,L to
A',B',C',L'.
Hence, P lies on LL'.
Friendly. Jean-Pierre
• Dear Alexey, ... (It is actually Lemoine , but better say symmedian points .) ... I have found an elementary proof of this fact, hoping that you will like
Message 3 of 9 , Jun 20, 2004
Dear Alexey,

In Hyacinthos message #9856, you wrote:

>> Let A'B'C' is the circumcevian triangle of
>> ABC with respect to point P, L, L' are the
>> Lemuan's point

(It is actually "Lemoine", but better say
"symmedian points".)

>> of ABC and A'B'C'. Then P is in LL'.

I have found an elementary proof of this fact,
hoping that you will like it.

The following lemma is important:

LEMMA 1. Let BCD and B'C'D' be two triangles such
that the lines BB', CC', DD' are concurent at a
point A. Let B1, C1, D1 be three points on the
lines CD, DB, BC such that the lines BB1, CC1,
DD1 concur at a point M. Finally, let the lines
AB1, AC1 and AD1 meet the lines C'D', D'B' and
B'C' at three points B1', C1', D1'. Then, the
lines B'B1', C'C1', D'D1' and AM concur.

Here are two proofs of this lemma:

PROJECTIVE PROOF.

Using projective transformations, Lemma 1 is
trivial: Undertake such a projective
transformation which maps the axis of perspective
of triangles BCD and B'C'D' to the line at
infinity. Then, the (new) triangles BCD and
B'C'D' are homothetic (since opposite sidelines
are parallel). The center of homothety is A.
Then, the homothety with center A mapping
triangle BCD to triangle B'C'D' must, of course,
map the points B1, C1, D1 to the points B1', C1',
D1', respectively; thus, the point M lying on the
lines BB1, CC1, DD1 is mapped to a point M1 lying
on the lines B'B1', C'C1', D'D1' and on the line
AM, of course (what follows from the properties
of a homothety). Hence we are done.

ELEMENTARY PROOF.

Since the triangles BCD and B'C'D' are
perspective, i. e. the lines BB', CC' and DD'
concur (at A), the Desargues theorem yields that
the points X = CD /\ C'D', Y = DB /\ D'B' and
Z = BC /\ B'C' are collinear, where the sign /\
means "intersection" (i. e. the intersection of
two lines g and h is denoted by g /\ h).

Now consider the triangles BB'Y and B1B1'X. Since
the points BB' /\ B1B1' = A, B'Y /\ B1'X = D' and
YB /\ XB1 = D are collinear, the Desargues
theorem shows that the lines BB1, B'B1' and YX
concur. In other words, the point of intersection
S of the lines BB1 and B'B1' lies on the line
XYZ. Similarly, the point of intersection T of
the lines DD1 and D'D1' lies on the line XYZ,
too.

Finally we apply the Desargues theorem to the
triangles DD'T and BB'S: Since the lines DB, D'B'
and TS concur (at Y), the points DD' /\ BB',
D'T /\ B'S and TD /\ SB are collinear. But
DD' /\ BB' = A, D'T /\ B'S = D'D1' /\ B'B1' and
TD /\ SB = DD1 /\ BB1 = M. Hence, the points A,
D'D1' /\ B'B1' and M are collinear. In other
words, the lines D'D1', B'B1' and AM concur.
Similarly, the lines D'D1', C'C1' and AM concur.
Hence, all four lines B'B1', C'C1', D'D1' and AM
concur at one point. Proof complete.

Okay, now consider our two triangles ABC and
A'B'C', and their respective tangential triangles
XYZ and X'Y'Z'. The symmedian point of a triangle
is the perspector of this triangle with its
tangential triangle; hence, the lines AX, BY, CZ
meet at the symmedian point L of triangle ABC,
and the lines A'X', B'Y', C'Z' meet at the
symmedian point L' of triangle A'B'C'.

It is easy to see that not only the lines AA',
BB', CC', but also the lines XX', YY', ZZ' pass
through the point P. In fact, the line XX' passes
through the poles X and X' of the lines BC and
B'C' with respect to the circumcircle of triangle
ABC, and hence it is the polar of the point of
intersection of the lines BC and B'C'. But from
the cyclic quadrilateral BCC'B', it follows that
the polar of the point of intersection of the
lines BC and B'C' passes through the point of
intersection of the lines BB' and CC', i. e.
through P. Hence, the line XX' passes through P,
and similarly for the lines YY' and ZZ'.

Now, consider the triangles XYZ and X'Y'Z'. The
lines XX', YY', ZZ' are concurrent at the point
P. The points A, B, C are three points on the
lines YZ, ZX, XY, and we know that the lines XA,
YB, ZC concur at the point L. Furthermore, the
lines PA, PB and PC meet the lines Y'Z', Z'X',
X'Y' at the points A', B', C' (since the points
A', B', C' lie on the lines PA, PB, PC and on the
lines Y'Z', Z'X', X'Y'). Hence, an application of
Lemma 1 yields that the lines X'A', Y'B', Z'C'
and PL concur. In other words, the point L' where
the lines X'A', Y'B', Z'C' meet must lie on the
line PL. Hence, the points P, L and L' are
collinear. Proof complete.

Your theorem has a very interesting special case:
Namely, let P be the orthocenter H of triangle
ABC. Then, the points A', B', C' are the
intersections of the altitudes of triangle ABC
with the circumcircle (different from A, B, C,
respectively), and, equivalently, the points A',
B', C' are the reflections of the orthocenter H
in the sides BC, CA, AB. Thus, if DEF is the
orthic triangle of the triangle ABC, then the
triangle A'B'C' is the image of the triangle DEF
in the homothety with center H and factor 2.
Hence, the symmedian point L' of triangle A'B'C'
is the image of the symmedian point L1 of
triangle DEF in this homothety. It follows that
the point L' lies on the line HL1. Now, since by
your theorem the points P ( = H ), L and L' are
collinear, it follows that all four points H, L,
L' and L1 are collinear. Now forget L' and
paraphrase the collinearity of the points H, L
and L1 as follows:

The orthocenter of a triangle, the symmedian
point of the triangle, and the symmedian point of
the orthic triangle are collinear.

Finally, if Ia, Ib, Ic are the excenters and I is
the incenter of triangle ABC, then the triangle
ABC is the orthic triangle of the triangle
IaIbIc. Also, the orthocenter of triangle IaIbIc
is the incenter I of triangle ABC, and the
symmedian point of triangle IaIbIc is the Mitten
point of triangle ABC (this is X(9) in
Kimberling's ETC). Hence, we summarize:

The incenter, the Mitten point and the symmedian
point of any triangle are collinear.

This fact is rather trivial if you use trilinear
coordinates, but I have not seen any synthetic
proof before. Thanks for your general result.

Friendly,
Darij Grinberg
• On 20-06-04, Darij Grinberg wrote ... In other words: Let ABC be a triangle, P a point, A B C the circumcevian triangle of P, and
Message 4 of 9 , Jun 20, 2004
On 20-06-04, "Darij Grinberg" <darij_grinberg@...> wrote
(partly):

>Dear Alexey,
>
>In Hyacinthos message #9856, you wrote:
>
>>> Let A'B'C' is the circumcevian triangle of
>>> ABC with respect to point P, L, L' are the
>>> Lemuan's point
>
>(It is actually "Lemoine", but better say
>"symmedian points".)
>
>>> of ABC and A'B'C'. Then P is in LL'.

In other words:

Let ABC be a triangle, P a point, A'B'C'
the circumcevian triangle of P, and K,K'
the Lemoine points of ABC, A'B'C', resp.
The locus of P such that P,K,K' are collinear
is the entire plane.

Now, how about if we replace Ks with other
triangle centers (O[: trivial case], H,I,G,N etc)

Which are the corresponding loci of P ?

APH

--
• Dear Antreas, ... I have tried P = X(i) with 1
Message 5 of 9 , Jun 20, 2004
Dear Antreas,

In Hyacinthos message #9924, you wrote:

>> Let ABC be a triangle, P a point, A'B'C'
>> the circumcevian triangle of P, and K,K'
>> the Lemoine points of ABC, A'B'C', resp.
>> The locus of P such that P,K,K' are collinear
>> is the entire plane.
>>
>> Now, how about if we replace Ks with other
>> triangle centers

I have tried P = X(i) with 1 <= i <= 10, and
the only i's such that the locus is the
entire plane were X(3) - trivial indeed - and
X(6).

What is rather interesting is the locus of all
P such that the orthocenters of triangles ABC
and A'B'C' are collinear with P. This locus
passes through X(1), X(3) and X(4), through
the vertices of the triangle ABC and through
the feet of its altitudes. The locus doesn't
pass through X(2), X(i) for 5 <= i <= 10 and
i = 13 and i = 15. However, note that the
circumcircle of triangle ABC is obviously part
of the locus, hence, after dividing away the
circumcircle, I don't know whether the
vertices of triangle ABC remain. Another part
of the locus is the line at infinity, what is
very easy to see, too. Maybe some specialist
- Bernard? - could provide more facts?

Sincerely,
Darij Grinberg
• Dear Antreas and Darij, ... your locus with Hs is the line at infinity and Q021. http://perso.wanadoo.fr/bernard.gibert/curves/q021.html I will add this to my
Message 6 of 9 , Jun 20, 2004
Dear Antreas and Darij,

> [APH] Let ABC be a triangle, P a point, A'B'C'
> >> the circumcevian triangle of P, and K,K'
> >> the Lemoine points of ABC, A'B'C', resp.
> >> The locus of P such that P,K,K' are collinear
> >> is the entire plane.
> >> Now, how about if we replace Ks with other
> >> triangle centers
>
> [DG] What is rather interesting is the locus of all
> P such that the orthocenters of triangles ABC
> and A'B'C' are collinear with P. This locus
> passes through X(1), X(3) and X(4), through
> the vertices of the triangle ABC and through
> the feet of its altitudes. The locus doesn't
> pass through X(2), X(i) for 5 <= i <= 10 and
> i = 13 and i = 15. However, note that the
> circumcircle of triangle ABC is obviously part
> of the locus, hence, after dividing away the
> circumcircle, I don't know whether the
> vertices of triangle ABC remain. Another part
> of the locus is the line at infinity, what is
> very easy to see, too. Maybe some specialist
> - Bernard? - could provide more facts?

your locus with Hs is the line at infinity and Q021.

I will add this to my web site.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Antreas, ... Hs : Q021 (my previous message) Gs : Q019 + Linf + C(O,R) Ns : Linf + a bicircular sextic which I plan to add to my web site... Best regards
Message 7 of 9 , Jun 20, 2004
Dear Antreas,

> [APH] Let ABC be a triangle, P a point, A'B'C'
> the circumcevian triangle of P, and K,K'
> the Lemoine points of ABC, A'B'C', resp.
> The locus of P such that P,K,K' are collinear
> is the entire plane.
>
> Now, how about if we replace Ks with other
> triangle centers (O[: trivial case], H,I,G,N etc)
>
> Which are the corresponding loci of P ?

Hs : Q021 (my previous message)

Gs : Q019 + Linf + C(O,R)

Ns : Linf + a bicircular sextic which I plan to add to my web site...

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, ... Note that all three points (H,G,N) are collinear. Probably we get interesting curves (as loci) for other points on the same line (Euler line)
Message 8 of 9 , Jun 20, 2004
Dear Bernard,

>> [APH] Let ABC be a triangle, P a point, A'B'C'
>> the circumcevian triangle of P, and K,K'
>> the Lemoine points of ABC, A'B'C', resp.
>> The locus of P such that P,K,K' are collinear
>> is the entire plane.
>>
>> Now, how about if we replace Ks with other
>> triangle centers (O[: trivial case], H,I,G,N etc)
>>
>> Which are the corresponding loci of P ?

[BG]:

>Hs : Q021 (my previous message)
>
>Gs : Q019 + Linf + C(O,R)
>
>Ns : Linf + a bicircular sextic which I plan to add to my web site...

Note that all three points (H,G,N) are collinear.

Probably we get interesting curves (as loci) for other points
on the same line (Euler line)

How about the Ls [de Longchamps], for example?

Greetings

Antreas

--
• Dear Antreas, ... Let X be a center on the Euler line such that OX = k OH (vectors, k¡Á0) and X the same center in A B C . P, X, X are collinear iff P lies
Message 9 of 9 , Jun 21, 2004
Dear Antreas,

> >> [APH] Let ABC be a triangle, P a point, A'B'C'
> >> the circumcevian triangle of P, and K,K'
> >> the Lemoine points of ABC, A'B'C', resp.
> >> The locus of P such that P,K,K' are collinear
> >> is the entire plane.
> >>
> >> Now, how about if we replace Ks with other
> >> triangle centers (O[: trivial case], H,I,G,N etc)
> >>
> >> Which are the corresponding loci of P ?
>
> [BG]:
>
> >Hs : Q021 (my previous message)
> >
> >Gs : Q019 + Linf + C(O,R)
> >
> >Ns : Linf + a bicircular sextic which I plan to add to my web site...
>
> Note that all three points (H,G,N) are collinear.
>
> Probably we get interesting curves (as loci) for other points
> on the same line (Euler line)
>
> How about the Ls [de Longchamps], for example?

Let X be a center on the Euler line such that OX = k OH (vectors, k≠0)
and X' the same center in A'B'C'.
P, X, X' are collinear iff P lies on a bicircular sextic passing
through A, B, C (double points), O, X1113, X1114, the centers of the
Apollonian circles, the cevians of X in ABC, two other points E1, E2 on
the Euler line and on the circle centered at X468 orthogonal to the
circumcircle. Its has two real asymptotes parallel to those of the
rectangular circum-hyperbola passing through Y such that OY = (1-k)/2
OH.
This sextic decomposes into the circumcircle and a circular quartic iff
X = G. This is Q019.

Best regards

Bernard

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.