- Dear Antreas,

>

labelling

> [APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three

> squares based on BC,CA,AB outwardly ABC.

>

> Ba = BC /\ AA2, Ca = BC /\ AA3, Similarly Cb, Ab; Bc, Ac.

> >

> >

> > \ /

> > Ab Ac

> > \ /

> > \ /

> > \ /

> > A3 \ / A2

> > \ /

> > A

> > /\

> > / \

> > / \

> > / \

> > / \

> > B3 / \ C2

> > / \

> > ---Ba------------------B--------------C------------------Ca-----

> > / \

> > / \

> > / \

> > / \

> > Bc Cb

> > / B1 C1 \

> >

> OLD PROBLEMS:

> ...

> [APH, Hyacinthos, message #2118, 18 Dec 2000]:

> Are the perpendicular bisectors of A2A3, B3B1, C1C2 concurrent?

>

> [FvL, Hyacinthos, message #2123, 19 Dec 2000]:

> These perpendicular bisectors are concurrent at:

> ( ... : 2SS + 2SSB - bbS + 4SAC + bbSB - aaSA - ccSC : ... )

> [in barycentrics]

>

> [APH]: NEW PROBLEMS:

> 1. Are the perpendicular bisectors of B3C2, C1A3, A2B1 concurrent?

> 2. Let A' = (perp. bis of BC) /\ (perp. bis. of A2A3)

> Similarly B', C'.

> Are the triangles ABC, A'B'C' perspective?

>

> ***

> I shall follow the notation in Floor's Friendship paper by

> these squares BA_bA_cC, CB_cB_aA, and AC_aC_bB [so that A2, A3 are

barycentric

> the points B_a, C_a respectively]. These have homogeneous

> coordinates

of

>

> B_a = (SC+S : -b^2 : S_A),

> C_a = (SB+S : S_A : -c^2).

>

> The midpoint of the segment is A' = (a^2+2S : -SC : -SB).

>

> Similarly,

> B_c = (SC : -b^2 : SA+S),

> C_b = (SB : SA+S : -c^2).

>

> The midpoint of the segment B_cC_b is A'' = (-a^2 : SC-S : SB-S),

> which is the center of the square erected on BC, on the same side

> A. This point is also equidistant from B_a and C_a.

perpendicular

>

> This gives an immediate solution to New Problem 2: the

> bisectors of BC and B_aC_a intersect at A''. It follows that

These coordinates [of the inner Vecten point] should of course be

> A''B''C'' is perspective with ABC at the inner Vecten point

>

> (SA-S : SB-S : SC-S).

(1/(SA-S) : 1/(SB-S) : 1/(SC-S))

instead.

Best regards

Sincerely

Paul - THIS MESSAGE CONTAINS HYACINTHOS #6445, CORRECTED AFTER HYACINTHOS

#6463, AND SOME NEW RESULTS.

We will treat two Grebe triangles of a triangle: the outer Grebe

triangle and the inner Grebe triangle.

--- THE OUTER GREBE TRIANGLE

Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of

triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the

lines BaCa, CbAb, AcBc is called the outer Grebe triangle of ABC. It

is perspective (and homothetic) with triangle ABC, the perspector

being the symmedian point of ABC (this is a well-known fact).

With the help of computer reducing, I have found out that the outer

Grebe triangle is also in perspective with the orthic triangle of

ABC. The perspector, which I call outer Grebe-orthic perspector, has

homogeneous trilinears

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \

( ---------------- : ---------------- : ---------------- ),

\ cos A cos B cos C /

where D is the area of triangle ABC.

By the way, for A' we have trilinears

A' ( 2D + b² + c² : -ab : -ca ),

or equivalently A' ( - 2D - b² - c² : ab : ca ).

The circumcenter T of the outer Grebe triangle A'B'C' lies on the

Brocard axis of ABC. The trilinears of T are

( (1 + 2 cot w) cos A - 2 sin A

: (1 + 2 cot w) cos B - 2 sin B

: (1 + 2 cot w) cos C - 2 sin C ),

where w is the Brocard angle of triangle of triangle ABC.

(Note that 1 + 2 cot w is the homothetic factor of triangles ABC and

A'B'C': for example, B'C' = (1 + 2 cot w) a.)

We can also write the trilinears of T as

( (2area + a² + b² + c²)/(2area) cos A - 2 sin A : ... )

= ( (1 + 2 cot A + 2 cot B + 2 cot C) cos A - 2 sin A : ... ).

Neither the outer Grebe-orthic perspector, nor T is (yet) in ETC.

--- THE INNER GREBE TRIANGLE

Edward Brisse had pointed me to a modification of the outer Grebe

triangle:

The inner Grebe triangle results if we erect the squares BBaCaC,

CCbAbA, AAcBcB inwards instead of outwards. Then, we get the inner

Grebe triangle A"B"C" enclosed by the lines BaCa, CbAb, AcBc.

This inner Grebe triangle A"B"C" is also perspective to the orthic

triangle of ABC. The perspector is called the inner Grebe-orthic

perspector and has trilinears

/ (D-a²)(2D-b²-c²) (D-b²)(2D-c²-a²) (D-c²)(2D-a²-b²) \

( ---------------- : ---------------- : ---------------- ).

\ cos A cos B cos C /

The vertices of the inner Grebe triangle have trilinears

A" ( 2D - b² - c² : ab : ca ) etc..

The circumcenter T' of the inner Grebe triangle A"B"C" lies on the

Brocard axis of ABC. The trilinears of T' are

( (1 - 2 cot w) cos A + 2 sin A

: (1 - 2 cot w) cos B + 2 sin B

: (1 - 2 cot w) cos C + 2 sin C ),

where w is the Brocard angle of triangle of triangle ABC.

(Now 1 - 2 cot w is the homothetic factor of triangles ABC and

A"B"C": for example, B"C" = (1 - 2 cot w) a.)

We can also write the trilinears of T' as

( (2area - a² - b² - c²)/(2area) cos A + 2 sin A : ... )

= ( (1 - 2 cot A - 2 cot B - 2 cot C) cos A + 2 sin A : ... ).

Neither the inner Grebe-orthic perspector, nor T' is (yet) in ETC.

--- TWO POINTS WITHOUT GEOMETRICAL DESCRIPTION

We have identified the points with trilinears

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \

( ---------------- : ---------------- : ---------------- )

\ cos A cos B cos C /

and

/ (D-a²)(2D-b²-c²) (D-b²)(2D-c²-a²) (D-c²)(2D-a²-b²) \

( ---------------- : ---------------- : ---------------- )

\ cos A cos B cos C /

as the outer and inner Grebe-orthic perspectors. But we can also

define two other points,

/ (D+a²)(2D-b²-c²) (D+b²)(2D-c²-a²) (D+c²)(2D-a²-b²) \

( ---------------- : ---------------- : ---------------- )

\ cos A cos B cos C /

and

/ (D-a²)(2D+b²+c²) (D-b²)(2D+c²+a²) (D-c²)(2D+a²+b²) \

( ---------------- : ---------------- : ---------------- ),

\ cos A cos B cos C /

which are also not in ETC. I don't know of a geometrical

signification of these points.

Darij Grinberg