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

Expand Messages
• An interesting thread was started in the geometry-college group: http://mathforum.org/epigone/geometry-college/pringkheldeh ... Subject: Brocard points of
Message 1 of 15 , Jun 1, 2004
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

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

----------------------------------------------------
• Darij raised some interesting questions about harmonic quadrilaterals. ... Casey, A sequel to Euclid , has various theorems on harmonic n-gons, including a
Message 2 of 15 , Jun 1, 2004

> 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.

Casey, 'A sequel to Euclid', has various theorems on harmonic n-gons,
including a formula for the Bocard angle ascribed to Tucker, namely:
cosec^w = cosec^A + cosec^B + ....

Casey also gives the result (for n=4) about equal products of opposite
sides. It is also in Johnson's Advanced Euclidean Geometry' (Th 133) which
also quotes (Th504) further Brocard properties ascribed to Tucker.

Dick Tahta
• Dear Dick, ... Thank you very much for the information. My dynamic sketch confirms csc^2 w = csc^2 A + csc^2 B for a harmonic quadrilateral. But what is the
Message 3 of 15 , Jun 1, 2004
Dear Dick,

In Hyacinthos message #9830, you wrote:

>> Casey, 'A sequel to Euclid', has various
>> theorems on harmonic n-gons, including a
>> formula for the Bocard angle ascribed to
>> Tucker, namely:
>> cosec^w = cosec^A + cosec^B + ....

Thank you very much for the information.
My dynamic sketch confirms

csc^2 w = csc^2 A + csc^2 B

for a harmonic quadrilateral. But what is
the definition of an harmonic n-gon?

Sincerely,
Darij Grinberg
• Dear Geoff Millin and Darij! ... This results are very interesting. I have one question. It is known that the projections of two isogonally conjugated points
Message 4 of 15 , Jun 2, 2004
Dear Geoff Millin and Darij!
>
>>> 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 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.
>
This results are very interesting. I have one question. It is known that the
projections of two isogonally conjugated points to the sidelines of ABCD are
in the circle. The center of this circle is the midpoint of segment RR' and
it is in Gauss line of ABCD. What is this center if R and R' are the Brocard
points?

Sincerely Alexey
• Dear Alexey, ... This is very interesting! At first, consider the Brocard points R and R of the harmonic quadrilateral ABCD, and the symmedian point K (this
Message 5 of 15 , Jun 2, 2004
Dear Alexey,

In Hyacinthos message #9842, you wrote:

>> It is known that the projections of two
>> isogonally conjugated points to the
>> sidelines of ABCD are in the circle. The
>> center of this circle is the midpoint of
>> segment RR' and it is in Gauss line of
>> ABCD. What is this center if R and R'
>> are the Brocard points?

This is very interesting! At first, consider
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

Would be very nice if anybody had a proof to
all of this...

Sincerely,
Darij Grinberg
• Dear Darij! ... I can prove some of this facts. Let P is the first Brocard point and the lines AP, BP, CP, DP intersect the circumcircle in B , C , D , A .
Message 6 of 15 , Jun 4, 2004
Dear Darij!
>
>This is very interesting! At first, consider
>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
>
>Would be very nice if anybody had a proof to
>all of this...
>
I can prove some of this facts. Let P is the first Brocard point and the
lines AP, BP, CP, DP intersect the circumcircle in B', C', D', A'. Then the
quadrilateral A'B'C'D' is the rotation of ABCD with center O and angle 2x,
where x is the Brocard angle. So P is the second Brocard point Q' of
A'B'C'D', OP=OQ and angle POQ=2x. Now let K is the common point of AC and
BD, K' - the common point of A'C' and B'D'. Then P is in KK' (see my paper
in "Kvant" N.1, 1996). Similarly Q is in KK'', where K'' is the reflection
of K' in OK. As K''Q=KP and POQ=K'OK=KOK''=2x, P and Q are the midpoints of
KK' and KK" respectively. So the angles OPK and OQK are direct and P, Q are
in the circle with diameter OK. Also it is easy to find x from equation
sin^4(x)=sin(A-x)sin(B-x)sin(C-x)sin(D-x), but I don't understand why ABCD
must be harmonic.

Sincerely Alexey
• Dear Darij! ... There is one correction. P is the midpoint of KK because OK=OK . This follows from next theorem. Let four chords A_iB_i concur in point C. D
Message 7 of 15 , Jun 4, 2004
Dear Darij!
>>
>>
>I can prove some of this facts. Let P is the first Brocard point and the
>lines AP, BP, CP, DP intersect the circumcircle in B', C', D', A'. Then the
>quadrilateral A'B'C'D' is the rotation of ABCD with center O and angle 2x,
>where x is the Brocard angle. So P is the second Brocard point Q' of
>A'B'C'D', OP=OQ and angle POQ=2x. Now let K is the common point of AC and
>BD, K' - the common point of A'C' and B'D'. Then P is in KK' (see my paper
>in "Kvant" N.1, 1996). Similarly Q is in KK'', where K'' is the reflection
>of K' in OK. As K''Q=KP and POQ=K'OK=KOK''=2x, P and Q are the midpoints of
>KK' and KK" respectively. So the angles OPK and OQK are direct and P, Q are
>in the circle with diameter OK.
There is one correction. P is the midpoint of KK' because OK=OK'. This
follows from next theorem. Let four chords A_iB_i concur in point C. D is
the common point of A_1A_2 and A_3A_4, E - the common point of B_1B_2 and
B_3B_4. Then C, D, E are collinear and in Klein's model of Lobachevsky
geometry C is the midpoint of DE.
>
Sincerely Alexey
• Dear colleagues! I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then 1) The Brocard points exist iff the symmedians of triangles
Message 8 of 15 , Jun 7, 2004
Dear colleagues!
I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
1) The Brocard points exist iff the symmedians of triangles
A_{i-1}A_iA_{i+1} from A_i concur in point L.
2) Two Brocard points P, Q are in the circle with diameter OL and the angles
POL and QOL are equal to brocard angle.
If 1) is correct then I can prove 2)

Sincerely Alexey
• Dear Alexey ... the angles ... And what do you think of the following conjecture : A cyclic polygon has Brocard points if and only if the polygon is the
Message 9 of 15 , Jun 8, 2004
Dear Alexey
> I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
> 1) The Brocard points exist iff the symmedians of triangles
> A_{i-1}A_iA_{i+1} from A_i concur in point L.
> 2) Two Brocard points P, Q are in the circle with diameter OL and
the angles
> POL and QOL are equal to brocard angle.
> If 1) is correct then I can prove 2)

And what do you think of the following conjecture :
A cyclic polygon has Brocard points if and only if the polygon is
the inverse of a regular polygon?
Friendly. Jean-Pierre
• Dear Jean-Pierre! ... I think that this two conditions are equivalent. If the point L exists then the projective transformation T conserving the circumcircle
Message 10 of 15 , Jun 8, 2004
Dear Jean-Pierre!
>> I have a next hypothesis. Let A_1...A_n is the cyclic polygon. Then
>> 1) The Brocard points exist iff the symmedians of triangles
>> A_{i-1}A_iA_{i+1} from A_i concur in point L.
>
>And what do you think of the following conjecture :
>A cyclic polygon has Brocard points if and only if the polygon is
>the inverse of a regular polygon?
>
I think that this two conditions are equivalent. If the point L exists then
the projective transformation T conserving the circumcircle and such that
T(L)=O transforms our polygon to regular. In the circle this transformation
coincide with some inversion.

My condition 1) follows from next fact. Let A, B, C, D are in the circle
with center O. P is the point such that the angles PAB, PBC and PCD are
equal, Q is the point such that the angles QDC, QCB and QBA are equal (P is
the common point distinct from B of two circles: one pass through A and B
and touches BC, and another pass through B and C and touches CD), L is the
common point of symmedians BX and CY of triangles ABC and BCD. Then O, P, Q,
L are in the circle.
It seems that this fact can be proved by calculation, but it is interesting
to find the synthetic proof.

Sincerely Alexey
• Dear Alexey [AZ] ... Then ... [JPE] ... [AZ] ... exists then ... such that ... transformation ... Suppose that A_1,...,A_n has Brocard points P,Q; let M be a
Message 11 of 15 , Jun 8, 2004
Dear Alexey
[AZ]
> >> I have a next hypothesis. Let A_1...A_n is the cyclic polygon.
Then
> >> 1) The Brocard points exist iff the symmedians of triangles
> >> A_{i-1}A_iA_{i+1} from A_i concur in point L.
[JPE]
> >And what do you think of the following conjecture :
> >A cyclic polygon has Brocard points if and only if the polygon is
> >the inverse of a regular polygon?
[AZ]
> I think that this two conditions are equivalent. If the point L
exists then
> the projective transformation T conserving the circumcircle and
such that
> T(L)=O transforms our polygon to regular. In the circle this
transformation
> coincide with some inversion.

Suppose that A_1,...,A_n has Brocard points P,Q; let M be a Poncelet
point of the pencil generated by the circumcircle (O) of the polygon
and the circle OPQ. the line MA_k intersects again (O) at B_k.
Then B_1,...,B_n is regular.
Of course, it is necessary to prove that such a point M is real, ie
that (O) and the circle OPQ cannot intersect.
It is necessary too to prove the reciprocal.
I think that it could be possible to name these two points M the
isodynamic points of the Brocardian polygon because
MA_(i+k)/M_A(i-k) = A_iA_(i+k)/A_iA_(i-k) for all i and k.
Friendly. Jean-Pierre
• ... Yes, this is the same transformation. Let we considere a sphere with diameter circle (O) and the pole N. The perpendicular to the plan (O) from my point L
Message 12 of 15 , Jun 9, 2004
>[AZ]
>> I think that this two conditions are equivalent. If the point L
>exists then
>> the projective transformation T conserving the circumcircle and
>such that
>> T(L)=O transforms our polygon to regular. In the circle this
>transformation
>> coincide with some inversion.
>
>Suppose that A_1,...,A_n has Brocard points P,Q; let M be a Poncelet
>point of the pencil generated by the circumcircle (O) of the polygon
>and the circle OPQ. the line MA_k intersects again (O) at B_k.
>Then B_1,...,B_n is regular.
>Of course, it is necessary to prove that such a point M is real, ie
>that (O) and the circle OPQ cannot intersect.
>It is necessary too to prove the reciprocal.
>I think that it could be possible to name these two points M the
>isodynamic points of the Brocardian polygon because
>MA_(i+k)/M_A(i-k) = A_iA_(i+k)/A_iA_(i-k) for all i and k.
>
Yes, this is the same transformation. Let we considere a sphere with
diameter circle (O) and the pole N. The perpendicular to the plan (O) from
my point L intersect the sphere in point L' (the plan (O) divide L' and N).
Then the line NL' intersect (O) in your point M. (That is the corresponding
between Klein and Poincare models of Lobachevsky geometry). So the inversion
with center M is equivalent to projective transformation with center L.

Sincerely Alexey
• Dear Alexey, suppose that ABC is a variable isosceles triangle (in B) inscribed in a fixed circle with center o; let M be a fixed point and A B C the cevian
Message 13 of 15 , Jun 9, 2004
Dear Alexey,
suppose that ABC is a variable isosceles triangle (in B) inscribed
in a fixed circle with center o; let M be a fixed point and A'B'C'
the cevian triangle of M wrt ABC. It is very easy to check that the
B'-symedian of A'B'C' intersects the line OM at a point L
independant of the choice of ABC.
If S is a point of the circle such as OM and OS are perpendicular, L
is the projection upon OM of the reflection of O wrt SM. More over L
lies on the half-line OM and inside the circle.
From this it follows that if B_1,...,B_n is a regular polygon
inscribed in the circle, if A_k is the second intersection of MB_k
with the circle, every A_k symedian of A_(k-1),A_k,A_(k+) goes
through L.
That's probably a part of the proof you're waiting for.
Friendly. Jean-Pierre
• Dear Alexey, I wrote ... Of course, I was meaning the circumcevian triangle of M wrt ABC. Sorry for the typo. ... L ... L
Message 14 of 15 , Jun 9, 2004
Dear Alexey,
I wrote
> suppose that ABC is a variable isosceles triangle (in B) inscribed
> in a fixed circle with center o; let M be a fixed point and A'B'C'
> the cevian triangle of M wrt ABC.

Of course, I was meaning the circumcevian triangle of M wrt ABC.
Sorry for the typo.

> It is very easy to check that the
> B'-symedian of A'B'C' intersects the line OM at a point L
> independant of the choice of ABC.
> If S is a point of the circle such as OM and OS are perpendicular,
L
> is the projection upon OM of the reflection of O wrt SM. More over
L
> lies on the half-line OM and inside the circle.
> From this it follows that if B_1,...,B_n is a regular polygon
> inscribed in the circle, if A_k is the second intersection of MB_k
> with the circle, every A_k symedian of A_(k-1),A_k,A_(k+) goes
> through L.
> That's probably a part of the proof you're waiting for.
> Friendly. Jean-Pierre
• ... These remarks started an interesting thread, which I could not follow since one half of the mathematics involved was new to me, but let me just make one
Message 15 of 15 , Jun 9, 2004
In Hyacinthos message #9843, I wrote:

>> This is very interesting! At first, consider
>> 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.

These remarks started an interesting thread,
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
Your message has been successfully submitted and would be delivered to recipients shortly.