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## Re: Orthial Triangles

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• ... Theorem #3: Theorem #3 The feet Ta, Tb, Tc, of the exsymmedians: ATa of the triangle ABaCa, BTb of the triangle BCbAb, and CTc of the triangle CBcAc, are
Message 1 of 7 , Dec 31, 1969
I wrote:

>Following are some theorems / conjectures on what I call "orthial triangles".
>
>Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
>A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :
>
> \ /
> Ab Ac
> \ /
> \ /
> \ /
> \ /
> \ /
> A
> /\
> / \
> Sc Mb
> / \
> Mc Sb
> / \
> / \
> ---Ba------------------B---Sa----Ma---C------------------Ca-----
> / \
> / \
> / \
> / \
> Bc Cb
> / \
>
>In the figure: BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;
>
>CbB _|_ BA , AbB _|_ BA (The symbol _|_ means "perpendicular to")
>
>I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
>(for the lack of a better name!).
>
>Theorem #1
>The medians: AMa of the triangle ABaCa, BMb of the triangle
>BCbAb, and CMc of the triangle CBcAc are concurrent.
>(Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)
>
>Note: John Conway and I have discussed it in geometry-college.
>I had calculated trigonometrically the a-Ceva ratio:
>
>BMa sinC (cosC - cosBcosA)
>---- = ----- * ----------------
>MaC sinB (cosB - cosCcosA)
>
>and Conway proved it using barycentrics.
>
> http://forum.swarthmore.edu/epigone/geom.college/styterben/
>
>The point of concurrency is the de Longchamps Point of ABC.
>(X_20 in Clark Kimberling's _TCCT_).
>
>Theorem #2
>The symmedians: ASa of the triangle ABaCa, BSb of the triangle
>BCbAb, and CSc of the triangle CBcAc are concurrent.
>
>Note:
>I calculated the a-Ceva ratio:
>
>BSa sinC (cosB - cosAcosC)
>---- = ---- * -----------------
>SCa sinB (cosC - cosAcosB)
>
>The point of concurrency is X_64 in _TCCT_.
>(There is no mention of a geometric property of this point in
>that entry. So, we have now one!)

Theorem #3:

Theorem #3
The feet Ta, Tb, Tc, of the exsymmedians: ATa of the triangle ABaCa, BTb
of the triangle BCbAb, and CTc of the triangle CBcAc, are collinear.

I proved it by Menelaus theorem.

The a-Menelaus ratio is:

BTa sinC (cosB + cosAcosC)
--- = ---- * -----------------
TaC sinB (cosC + cosAcosB)

Conjectures:

I have drawn some figures with ISOPTIKON, and "discovered" that the
the perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

Also, the Euler lines of the four triangles ABC, ABaCa, BCbAb, CBcAc
are concurrent. (This point is probably an new point on the Euler line)

I didn't draw figures for other lines (OK, OI, etc), but I believe
that they concur as well.

Happy Holidays

Antreas
• ABC right-angled triangle at A sin^2(A) = sin^2(B) + sin^2(C) (Pythagorean Theorem) How about to call a triangle ABC as pseudo-right-angled if holds the
Message 2 of 7 , Dec 31, 1969
ABC right-angled triangle at A <==> sin^2(A) = sin^2(B) + sin^2(C)
(Pythagorean Theorem)

How about to call a triangle ABC as "pseudo-right-angled" if holds the
equality: cos^2(A) = cos^2(B) + cos^2(C) ?

Which might be a geometric property of such a triangle ?

Well... This one:
The bases of its "orthial triangles" form a right-angled triangle.

I wrote:

>Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
>A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :
>
> \ /
> Ab Ac
> \ /
> \ /
> \ /
> \ /
> \ /
> A
> /\
> / \
> Sc Mb
> / \
> Mc Sb
> / \
> / \
> ---Ba------------------B---Sa----Ma---C------------------Ca-----
> / \
> / \
> / \
> / \
> Bc Cb
> / \
>
>In the figure: BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;
>
>CbB _|_ BA , AbB _|_ BA (The symbol _|_ means "perpendicular to")
>
>I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
>(for the lack of a better name!).

The "bases" BaCa, CbAb, BcAc of these triangles are equal to:

atanBtanC, btanCtanA, ctanAtanB.

Now, if the triangle formed with these "bases" is right-angled (with the angle
opposite to BaCa equal to 90 d.) then:

(BaCa)^2 = (CbAb)^2 + (BcAc)^2 <==> (atanBtanC)^2 = (btanCtanA)^2+(ctanAtanB)^2
<==> cos^2(A) = cos^2(B) + cos^2(C)

Happy Holidays

Antreas
• Dear Friends, Seasonal Greetings from Athens. And Happy Birthday, John! Following are some theorems / conjectures on what I call orthial triangles . Let ABC
Message 3 of 7 , Dec 31, 1969
Dear Friends,

Seasonal Greetings from Athens.
And Happy Birthday, John!

Following are some theorems / conjectures on what I call "orthial triangles".

Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :

\ /
Ab Ac
\ /
\ /
\ /
\ /
\ /
A
/\
/ \
Sc Mb
/ \
Mc Sb
/ \
/ \
---Ba------------------B---Sa----Ma---C------------------Ca-----
/ \
/ \
/ \
/ \
Bc Cb
/ \

In the figure: BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;

CbB _|_ BA , AbB _|_ BA (The symbol _|_ means "perpendicular to")

I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
(for the lack of a better name!).

Theorem #1
The medians: AMa of the triangle ABaCa, BMb of the triangle
BCbAb, and CMc of the triangle CBcAc are concurrent.
(Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)

Note: John Conway and I have discussed it in geometry-college.
I had calculated trigonometrically the a-Ceva ratio:

BMa sinC (cosC - cosBcosA)
---- = ----- * ----------------
MaC sinB (cosB - cosCcosA)

and Conway proved it using barycentrics.

http://forum.swarthmore.edu/epigone/geom.college/styterben/

The point of concurrency is the de Longchamps Point of ABC.
(X_20 in Clark Kimberling's _TCCT_).

Theorem #2
The symmedians: ASa of the triangle ABaCa, BSb of the triangle
BCbAb, and CSc of the triangle CBcAc are concurrent.

Note:
I calculated the a-Ceva ratio:

BSa sinC (cosB - cosAcosC)
---- = ---- * -----------------
SCa sinB (cosC - cosAcosB)

The point of concurrency is X_64 in _TCCT_.
(There is no mention of a geometric property of this point in
that entry. So, we have now one!)

Conjecture:
The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
How about the Brocard Axes of the orthial triangles ABaCa, BCbAb, CBcAc?
I would bet that they concur too!

Antreas
• Many happy returns of the day, John. When will you next survive three consecutive orders of simple groups? R.
Message 4 of 7 , Dec 26, 1999
Many happy returns of the day, John.

When will you next survive three consecutive orders of simple groups? R.
• ... That s almost as bad as telling Erdos he was duplicating the cube for the last time! Looking forward to your next biquadrate, John C. PS. I never saw
Message 5 of 7 , Dec 27, 1999
On Sun, 26 Dec 1999, Richard Guy wrote:

> Many happy returns of the day, John.
>
> When will you next survive three consecutive orders of simple groups? R.

That's almost as bad as telling Erdos he was duplicating the cube
for the last time!

Looking forward to your next biquadrate, John C.

PS. I never saw that "Caian". What did it say? JHC
• ... Thanks! ... which should be enough for me to keep the notation in mind *** ... *** The two methods are essentially the same, since the Cevian ratios of
Message 6 of 7 , Dec 27, 1999
On Sat, 25 Dec 1999, Antreas P. Hatzipolakis wrote:

> Seasonal Greetings from Athens.
> And Happy Birthday, John!

Thanks!

> Following are some theorems / conjectures on what I call "orthial triangles".
>
> Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
> A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :

*** Note by JHC: I retain only this much of the figure:

> / \
> ---Ba------------------B---Sa----Ma---C------------------Ca-----
> / \

which should be enough for me to keep the notation in mind ***
> Theorem #1
> The medians: AMa of the triangle ABaCa, BMb of the triangle
> BCbAb, and CMc of the triangle CBcAc concur in L = deLongchamps pt.
> (Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)

> I had calculated trigonometrically the a-Ceva ratio:
>
> BMa sinC (cosC - cosBcosA)
> ---- = ----- * ----------------
> MaC sinB (cosB - cosCcosA)
>
> and Conway proved it using barycentrics.

*** The two methods are essentially the same, since the Cevian
ratios of the point with barycentrics (X:Y:Z) are Y:Z, Z:X, X:Y.
So the barycentrics of this point are ( : b(cosB - cosCcosA) : ),
since the sine law permits me to replace sinB by b. Alternatively,
it might be 1/that if your Cevian ratio is the other way up - let me
check: in my notation, SB = ca.cosB = (cc+aa-bb)/2, so multiplying
the above by abc we get ( : bbSB - SCSA : ) = ( : SASB+SBSC-SCSA : ),
which is indeed the superior of the orthocenter ( : SCSA : ) ***
> Theorem #2
> The symmedians: ASa of the triangle ABaCa, BSb of the triangle
> BCbAb, and CSc of the triangle CBcAc are concurrent.
>
>
> I calculated the a-Ceva ratio:
>
> BSa sinC (cosB - cosAcosC)
> ---- = ---- * -----------------
> SCa sinB (cosC - cosAcosB)

*** This is equivalent to barycentrics ( : b/(cosB - cosCcosA) : ),
that is to say, the conjugal, L*, of the deLongchamps point L ***

> (There is no mention of a geometric property of this point in
> that entry. So, we have now one!)
>
> Conjecture:
> The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
> How about the Brocard Axes of the orthial triangles ABaCa, BCbAb, CBcAc?
> I would bet that they concur too!

OK, let me work on these (about which I'm rather doubtful). I'll
draw the triangle my usual way round:

Cb-------C--------A-------Ab

The vectors _|_ a,b,c are (-aa:SC:SB),(SC:-bb:SA),(SB:SA:-cc),
so Cb = (SB:0:-cc), Ab = (-aa:0:SB), whose coordinate-sums are
-SA and -SC. I'll therefore multiply them by -SC and -SA
respectively, to get a common denominator:

Cb = (-SBSC:0:ccSC) Ab = (aaSA:0:SASB)

so their midpoint is (aaSA-SBSC:0:ccSC-SASB), whence the general
point on their perpendicular is this + some multiple of (SC:-bb:SA).

This makes it clear that these perpendiculars don't concur, since there's
no way to adjust these multiples to give a symmetric point (I think - I'll
check properly in a moment). But taking the multiplier SB gives us
the interesting point (aaSA:-bbSB:ccSC), which is one of the harmonic
associates of the circumcenter O = (:bbSB:). So at least we get a
theorem: these perpendiculars pass through the preCevian or harmonic
triangle of the circumcenter.

But I'm getting a thought which makes me think that perhaps they DO
concur, at a very interesting point. No - I think it's the wrong way
round. What I'm thinking of is the fact that the contact points of
the Steiner deltoid with the edges have the property that they are
the Cevian feet of one point and the pedal feet of another. But in
fact they're the pedal feet of L and the Cevian feet of the point I
used to call M but think I now call it the Retrocenter R.

I tried doing that check online, but it began to look a bit complicated,
so I'll postpone it.

JHC
• Antreas - I shorten so that I can see all your points at once: ... So this is the line (p:q:r|0) where q or 1/q = (cosB + cosCcosA)/sinB The fire alarm s
Message 7 of 7 , Dec 27, 1999
Antreas - I shorten so that I can see all your points at once:

On Mon, 27 Dec 1999, Antreas P. Hatzipolakis wrote:

> >The medians: AMa of the triangle ABaCa, BMb of the triangle
> >BCbAb, and CMc of the triangle CBcAc are concurrent.

*** at L = ( : sinB(cosB - cosCcosA): ) = (:bbSB-SCSA:) ***

> >The symmedians: ASa of the triangle ABaCa, BSb of the triangle
> >BCbAb, and CSc of the triangle CBcAc are concurrent.

*** at L* = (:bb/(bbSB-SCSA):)

> The feet Ta, Tb, Tc, of the exsymmedians: ATa of the triangle ABaCa, BTb
> of the triangle BCbAb, and CTc of the triangle CBcAc, are collinear.
>
> The a-Menelaus ratio is:
>
> BTa sinC (cosB + cosAcosC)
> --- = ---- * -----------------
> TaC sinB (cosC + cosAcosB)

So this is the line (p:q:r|0) where

q or 1/q = (cosB + cosCcosA)/sinB

The fire alarm's just gone off in this building, so I'll have
to leave. (It regularly does so, but one should obey!) JHC

>
>
> Conjectures:
>
> I have drawn some figures with ISOPTIKON, and "discovered" that the
> the perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
>
> Also, the Euler lines of the four triangles ABC, ABaCa, BCbAb, CBcAc
> are concurrent. (This point is probably an new point on the Euler line)
>
> I didn't draw figures for other lines (OK, OI, etc), but I believe
> that they concur as well.
>
>
> Happy Holidays
>
>
> Antreas
>
> >
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