## 9824The Brocard points of a quadrilateral (Geoff Millin)

Expand Messages
• Jun 1, 2004
• 0 Attachment
An interesting thread was started in the
geometry-college group:

http://mathforum.org/epigone/geometry-college/pringkheldeh

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

Author: geoff millin <cgmillin*tiscali.co.uk>
replace the sign * by @

Correction:

ABCD is a cyclic quadrilateral. I have proved
that a necessary and sufficient condition for
it to have Brocard points is that the products
of the opposite sides are equal. Ie

I expect this is a well-known fact, but I can
find no reference to it on the web.

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

Subject: Re: Brocard points of quadrilaterals
Author: Darij Grinberg

Dear Geoff Millin,

You wrote:

>> ABCD is a cyclic quadrilateral. I have proved
>> that a necessary and sufficient condition for
>> it to have Brocard points is that the products
>> of the opposite sides are equal. Ie

Congratulations for having discovered a nice new
fact in Euclidean Geometry! A cyclic
is called a "harmonic quadrilateral"; such
quadrilaterals were studied in the 19th century,
and many properties of them were uncovered. Just
two examples: The line AC is a symmedian in the
triangles DAB and BCD simultaneously, and the
line BD is a symmedian in the triangles ABC and
CDA simultaneously. And: There exists an
inversion mapping the points A, B, C, D to the
vertices of a square. But I have never seen your
result about Brocard points mentioned anywhere,
it is most likely new.

I have tried to explore your Brocard points by
dynamic geometry software. One of the nice
properties (I have no proof yet) of these points
is that the two "Brocard angles" are equal, much
like in a triangle. In other words, if you take
a harmonic quadrilateral ABCD and its two
Brocard points R and R', with

< RAB = < RBC = < RCD = < RDA and
< R'BA = < R'CB = < R'DC = < R'AD,

then

< RAB = < RBC = < RCD = < RDA
= < R'BA = < R'CB = < R'DC = < R'AD.

Equivalently, the Brocard points R and R' are
"isogonal conjugates" with respect to the
quadrilateral ABCD, i. e. the lines AR and AR'
are symmetric with respect to the angle
bisector of the angle DAB, and similarly for
the other pairs of lines. Generally, not every
point in the plane of our quadrilateral ABCD
has an isogonal conjugate, but these Brocard
points have isogonal conjugates, namely the
isogonal conjugate of one Brocard point is the
other one.

I don't see, however, any formula of the kind
cot w = cot A + cot B + cot C for the Brocard