## Altimedial Triangles

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• 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
Message 1 of 4 , Jan 30, 2000
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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
• ... 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
Message 2 of 4 , Jan 30, 2000
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I wrote:

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

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
• ... We have: 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
Message 3 of 4 , Feb 1, 2000
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I wrote:

>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) ::)

We have:

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) :
Message 4 of 4 , Feb 1, 2000
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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@...]
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:
>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) ::)

We have:

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