- Definition:

Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle

of a triangle ABC is named after, namely:

M_i : the midpoints of sides BC,CA,AB, resp.

H_i : the feet of the altitudes from A,B,C, resp.

E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)

(i = 1,2,3)

I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3

The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three

altimedials are congruent (and similar to ABC).

Theorem:

Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles

H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are

concurrent.

Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

Proof:

Lemma:

Let ABA'C be a quadrilateral, with sides AB = c, BA' = c', A'C = b', CA = b,

and angles CAA' = A_1, BAA' = A_2, ABC = B, ACB = C, A'BC = w (omega),

A'CB = q (theta).Then

sinA_1 b' sin(C+q)

------ = ---- * ---------

sinA_2 c' sin(B+w)

A

/\

/ \

c b

/ \

/ \

B---------C

\ w q /

\ /

c' b'

\ /

\/

A'

b' AA'

Triangle ACA': -------- = ------- (1)

sinA_1 sin(C+q)

c' AA'

Triangle ABA': -------- = ------- (2)

sinA_2 sin(B+w)

sinA_1 b' sin(C+q)

(1) & (2) ===> ------ = ---- * ---------

sinA_2 c' sin(B+w)

A

/\

/ \

/ \

/ \

/ \

/ \

/ \

M_3 M_2

/ \

/ \

/ I_1 \

/ \

/ \

/ \

B-------------------H_1------C

In the quadrilateral AM_3I_1M_2 we have:

angles AM_2M_3 = C = H_1M_2M_3, AM_3M_2 = B = H_1M_3M_2

angles I_1M_3M_2 = B/2, I_1M_2M_3 = C/2

AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'sin(B/2), M_2I_1 = r'sin(C/2)

[r' = inradius of H_1M_2M_3 = r/2]

Denote angles I_1AM_2 := A_1, I_1AM_3 := A_2.

By ap[lyimg the lemma above we get:

sinA_1 I_1M_2 sin(C+ C/2) sin(B/2)sin(3C/2)

------ = -------- * ------------ = -------------------

sinA_2 I_1M_3 sin(B+ B/2) sin(C/2)sin(3B/2)

Similarly, for the incenters I_2, I_3 we get:

sinB_1 sin(C/2)sin(3A/2)

------ = -------------------

sinB_2 sin(A/2)sin(3C/2)

sinC_1 sin(A/2)sin(3B/2)

------ = -------------------

sinC_2 sin(B/2)sin(3A/2)

Therefore the lines AI_1, BI_2, CI_3 concur. (Trig. version of Ceva Theorem).

Note:

From the above Ceva ratia we get the trilinears of the point of concurrence:

(x:y:z) = (sin(A/2)sin(3B/2)sin(3C/2) ::)

Locus Problem:

Let ABC be a triangle, A'B'C' the pedal triangle of a point P, and

M_1M_2M_3 the medial triangle of ABC, and I_1, I_2, I_3 the incenters of

the triangles A'M_2M_3, M_1B'M_3, M_1M_2C', respectively.

Which is the locus of P such that the lines AI_1, BI_2, CI_3 concur?

(In the thoerem above P = H).

The same for the circumcenters instead of the incenters

(the three circumcenters above coincide with N, the the center of the 9pc)

Antreas - I wrote:

>AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'sin(B/2), M_2I_1 = r'sin(C/2)

Read instead:

AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'/sin(B/2), M_2I_1 = r'/sin(C/2)

Sorry for the typos.

Antreas - I wrote:

>Definition:

We have:

>Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle

>of a triangle ABC is named after, namely:

>

>M_i : the midpoints of sides BC,CA,AB, resp.

>H_i : the feet of the altitudes from A,B,C, resp.

>E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)

>(i = 1,2,3)

>

>I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3

>The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three

>altimedials are congruent (and similar to ABC).

>

>Theorem:

>Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles

>H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are

>concurrent.

>

>Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

sin(A/2)sin(3B/2)sin(3C/2) :: =

sin(A/2)*[3sin(B/2) - 4sin^2(B/2)]*[3sin(C/2) - 4sin^2(C/2)] :: =

(3 - 4sin(B/2))*(3 - 4sin(C/2)) ::

Can it be further reduced ?

Antreas - Dear Anteas,

This perspector of I_1I_2I_3 and ABC is the point X_{79} in

Kimberling's list. As such, its TRILINEAR coordinates can be

written as

1/(1+2cos A) : 1/(1+2cos B) : 1/(1+2cos C).

Here is the derivation of the homogeneous BARYCENTRIC

coordinates.

*************

The incenter I_1 of triangle H_1M_2M_3 is the reflection of the Spieker point

b+c:c+a:a+b

in the perpendicular bisector of the segment M_2M_3. This line has equation

(b^2-c^2)x + (2a^2-b^2+c^2)y - (2a^2+b^2-c^2)z = 0.

From this The incenter I_1 has coordinates

a(b+c) : a^2+ab+b^2-c^2 : a^2-b^2+c^2+ac

= *** : 1/(c^2+ca+a^2-b^2) : 1/(a^2+ab+b^2-c^2).

Similarly, we write down the coordinates of I_2

and I_3. It is evident then that I_1I_2I_3 and

ABC are perspective at

1/(b^2+bc+c^2-a^2) : 1/(c^2+ca+a^2-b^2) :

1/(a^2+ab+b^2-c^2).

As pointed above, this is X_{79} in Kimberling's list.

Best regards.

Sincerely,

Paul

----------

From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]

Reply To: Hyacinthos@onelist.com

Sent: Tuesday, February 01, 2000 2:37 PM

To: Hyacinthos@onelist.com

Subject: Re: [EMHL] Altimedial Triangles

From: xpolakis@... (Antreas P. Hatzipolakis)

I wrote:

>Definition:

We have:

>Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle

>of a triangle ABC is named after, namely:

>

>M_i : the midpoints of sides BC,CA,AB, resp.

>H_i : the feet of the altitudes from A,B,C, resp.

>E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)

>(i = 1,2,3)

>

>I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3

>The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three

>altimedials are congruent (and similar to ABC).

>

>Theorem:

>Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles

>H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are

>concurrent.

>

>Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

sin(A/2)sin(3B/2)sin(3C/2) :: =

sin(A/2)*[3sin(B/2) - 4sin^2(B/2)]*[3sin(C/2) - 4sin^2(C/2)] :: =

(3 - 4sin(B/2))*(3 - 4sin(C/2)) ::

Can it be further reduced ?

Antreas

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