- An interesting thread was started in the

geometry-college group:

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

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Subject: Brocard points of quadrilaterals

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

AB.CD = AD.BC.

I expect this is a well-known fact, but I can

find no reference to it on the web.

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Subject: Re: Brocard points of quadrilaterals

Author: Darij Grinberg

Dear Geoff Millin,

You wrote:

>> ABCD is a cyclic quadrilateral. I have proved

Congratulations for having discovered a nice new

>> that a necessary and sufficient condition for

>> it to have Brocard points is that the products

>> of the opposite sides are equal. Ie

>> AB.CD = AD.BC.

fact in Euclidean Geometry! A cyclic

quadrilateral ABCD satisfying AB * CD = AD * BC

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

angle of a harmonic quadrilateral.

May I ask you how you proved your original

result? Did you perform a calculation using

coordinates or do you have a synthetic proof?

Do you also have an idea about my above

conjectures?

Sincerely,

Darij Grinberg

---------------------------------------------------- - In Hyacinthos message #9843, I wrote:

>> This is very interesting! At first, consider

These remarks started an interesting thread,

>> the Brocard points R and R' of the harmonic

>> quadrilateral ABCD, and the symmedian point

>> K (this is the point whose distances to the

>> sides of the quadrilateral are proportional

>> to the lengths of the sides). Also consider

>> the circumcenter O of our quadrilateral.

>> Then, the points O, K, R and R' lie on one

>> circle, which is analogous to the "Brocard

>> circle" of a triangle. The points O and K

>> are the endpoints of a diameter of this

>> circle. The points R and R' lie

>> symmetrically with respect to this diameter.

>> The circle through the points O, K, R and R'

>> also passes through the midpoints of the

>> diagonals AC and BD. The line joining these

>> midpoints meets the diameter OK at the

>> midpoint of the segment RR'. The angles

>> < ROK and < R'OK are equal to the Brocard

>> angle of the quadrilateral ABCD.

which I could not follow since one half of the

mathematics involved was new to me, but let me

just make one remark: In a harmonic

quadrilateral ABCD, the point K is the point

of intersection of the diagonals AC and BD.

[This is quite easy to prove.] Now, since the

midpoints M and N of the diagonals AC and BD

are the orthogonal projections of the

circumcenter O on these diagonals, the angles

< OMK and < ONK are right angles. Hence, M and

N lie on the circle with diameter OK.

Darij Grinberg