- I wrote:

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

Theorem #3:

>

>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.

>

>See the thread:

> 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

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 - 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

>

> >