- Richard Guy wrote:

>Perhaps `Zeman' ??? R.

No. It is ZEEMAN.

Julio's message in HM (Fri, 03 Dec 1999) on Z. reads:

_____________________________________

<quote>

Dear John, Antreas, & all,

1. If a, b, c, d denote four straight lines (of the same plane) such

that $d$ be parallel to the Euler line of the triangle $abc$, then

the three Euler lines

a' of bcd, b' of cda, c' of dab,

are parallel to a, b, c, respectively.

This theorem of professor P. Zeeman Gz [from Delft, renowned all over

the world as the city of Delft blue earthenware, in the Netherlands] was

published early in 1903 in the Dutch journal _Wiskundige Opgaven_ [vol

VIII (1899-1902), p 305]. I first heard of this theorem thanks to a brief

note [_Mathesis_ (3) 3 (1903) p 60] written by the famous triangle guru

Joseph Neuberg (1840-1926) of Luxembourg:

"Si quatre droites a, b, c, d d'un plan sont telles

que l'une d'elles est parall\ele \a la droite d'Euler

du triangle des trois autres, cette propri/et/e a

encore lieu pour chacune des autres droites."

Neuberg noticed he could easily prove this proposition by using a formula

he had discussed in an earlier volume of _Wiskundige Opgaven_ [vol III

(1886-1889), p 372].

2. An interesting complement to Zeeman's theorem is that of professor

C.A. Cikot [from Bois-le-Duc (s'Hertogenbosch) capital of North Brabant

province, south central Netherlands], a proposition first published in

an article written also by Neuberg [_Mathesis_ (3) 8 (1908) p 233]:

Let a, b, c, d denote four straight lines (of the same plane). If $d$

is the Euler line of the triangle $abc$, then the three Euler lines a',

b', c' respectively of the triangles $bcd$, $cda$, $dab$ form a triangle

congruent to $abc$, and the Euler line d' of $a'b'c'$ coincides with $d$.

What about "The Cikot Point"? ...

John, this theorem implicitly includes the congruence-by-reflection

statement you've independently discovered. Do you agree? ...

By the way, many thanks for your kindest words in your previous letter

to me on this thread!

With best regards,

Julio Gonzalez Cabillon

</quote>

I don't know what the initial P. stands for

but the name appears as:

ZEEMAN GZ., P.

ZEEMAN GZN, P.

Probably the P. stands for PIETER (must

not be confused with the 1902 Physics nobelist

PIETER ZEEMAN)

P. ZEEMAN GZN wrote several geometrical articles

in Dutch periodicals.

In my next message I will give the German abstract

of an article of his about triangle cubics.

Antreas

------------------------------------------------------------------------

-- - AntreasLocus: For which P's the L11, L22, L33 are concurrent?Homothetic center in terms of the coordinates of P?We have that ABC, triangle bounded by (L11, L22, L33) areof the triangle bounded by (L3, PA, PB).Let L3 be the Euler line of PAB and L33 the Euler lineandof the triangle bounded by (L2, PC, PA)Let L2 be the Euler line of PCA and L22 the Euler lineSimilarly:of the triangle bounded by (PB,PC,L1).Let L1 be the Euler line of PBC and L11 the Euler lineNow, let ABC be a triangle and P a point.homothetic to ABC. To homothetic center is calledSimilarly Lb, Lc. So the triangle bounded by (La,Lb,Lc) isis parallel to BCThe Euler line La of the triangle bounded by (AB,AC, L)Let ABC be a triangle and L its Euler linr.Gossard Theorem:The following is an obvious consequence of Gossard theorem,so probably was already studied.

Zeeman - Gossard perspector X(402)

homothetic.