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.

(See also the MathLinks thread

http://mathlinks.ro/viewtopic.php?t=5117 )

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