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## Re: [EMHL] Re: new centers/items in complete quadrangle and complete quadrilateral

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• Dear Angel and Chris, Another construction for this: Let Qi be the cyclocevian conjugate of Pi wrt PjPkPl.  QiQjQkQl could be called the Cyclocevian
Message 1 of 12 , Jun 29, 2012
Dear Angel and Chris,

Another construction for this: Let Qi be the cyclocevian conjugate of Pi wrt PjPkPl.  QiQjQkQl could be called the 'Cyclocevian Quadrangle' of PiPjPkPl, and its perspector is this new center, which might be called the 'Cyclocevian Center'.

A related center would be the centroid of the Cyclocevian Quadrangle.  Coordinates?

Best regards,

Randy Hutson

>________________________________
> From: Angel <amontes1949@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Friday, June 29, 2012 12:04 PM
>Subject: [EMHL] Re: new centers/items in complete quadrangle and complete quadrilateral
>
>

>Dear Chris van Tienhoven,
>
>I have a Quadrangle Center that is not currently in EQF.
>
>Let P1, P2, P3, P4 be the defining Quadrangle Points.
>Let S1 = P1.P4 /\ P2.P3, S2 = P1.P3 /\ P2.P4 and S3 = P1.P2/\ P3.P4.
>Now S1 S2 S3 is the QA-Diagonal Triangle of the Reference Quadrangle.
>
>For each vertex Pi, we take the triangle TjTkTl, where Tj the
>intersection of the sidelinea PkPl with circumcircle of the triangle S1S2S3 (other than S1, S2, S3)
>Qi = Perspector of the triangle PjPkPl and TjTkTl
>
>The "unknown" Quadrangle Center is the common intersection point of lines Pi.Qi
>
>1st CT-Coordinate:
>
>p(q+r)(p+2q+r)^2(p+q+2r)^2
>((1/(p+2q+r)^2)((p+r)(-c^4(p+r)^2(q+r)^2+
>(p+q)^2(a^4(p+r)^2+b^4(q+r)^2))
>((q+r)^2(2p+q+r)^2SA+(p+r)^2(p+2q+r)^2SB+
>(p-q)^2(p+q)^2SC))+
>(1/(p+q+2r)^2)((p+q)(c^4(p+r)^2(q+r)^2+(p+q)^2(a^4(p+r)^2-
>b^4(q+r)^2))((q+r)^2(2p+q+r)^2SA+
>(p-r)^2(p+r)^2SB+(p+q)^2(p+q+2r)^2SC)))
>
>1st DT-Coordinate: a^2(c^4p^2q^2 + (b^4p^2 - a^4q^2)r^2)
>
>Construction GeoGebra:
>
>http://amontes.webs.ull.es/geogebra/EQF_QA_Pn.html
>
>If ABC is the diagonal triangle of the Quadrangle with a vertex in triangle center X(n), Then the Quadriangle Center obtained here is the TCC-perspector of X(n). See the note just before X(1601) in ETC.
>
>Best regards,
>
>Angel Montesdeoca
>
>--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>>
>> Dear Friends,
>>
>> I got some very nice remarks following my announcement of the Encyclopedia of Quadri-Figures.
>> Seiichi Kirikami attended me on a special property of a Complete Quadrangle.
>> The Orthopoles of a line with respect to the four component triangles of any complete quadrangle lie on a straight line known as the Orthopolar Line of for the given complete quadrangle (see also "Orthopolar line" in Mathworld).
>> Definition Orthopole of a random line wrt some Triangle:
>> If perpendiculars are dropped on any line from the vertices of a triangle, then the perpendiculars to the opposite sides from their perpendicular feet are concurrent at a point called the Orthopole (see also "Orthopole" in Mathworld).
>>
>> I tried some things with the Orthopolar Line and found these results:
>> 1. The 4 Orthopoles of a line wrt the Component Triangles in a Complete Quadrangle are collinear. Let's name this the QA-Orthopole Line.
>> 2. The 4 Orthopoles of a line wrt the Component Triangles in a Complete Quadrilateral are also collinear. Let's name this the QL-Orthopole Line.
>> 3. Every QA-Orthopole Line of a line through QA-P4 (Isogonal Center) is a line through QA-P2 (Euler-Poncelet Point).
>> It looks like this is the only case that a pencil of lines through a point is being transformed into another pencil through a point.
>> The locus of their mutual crosspoint is a hyperbola through QA-P2 and QA-P4.
>> Its conic center is the midpoint of QA-P2 and QA-P4.
>> Both lines are parallel when they are parallel to the asymptotes of this hyperbola.
>> 4. A QL-Orthopole Line is always a line perpendicular to the Newton Line (QL-L1). Most Special.
>>
>> I refer here to some points as coded in the Encyclopedia of Quadri-Figures:
>> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>>
>> Are there any (synthetic) proofs for 1., 2., 3. and 4. ?
>> Are there any more related properties known?
>>
>> Best regards,
>>
>> Chris van Tienhoven
>>
>>
>> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@> wrote:
>>
>> > You can find this Encyclopedia of Quadri-Figures (EQF) at:
>> > http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>> >
>> > The results also can be downloaded in PDF-format at:
>> >
>>
>
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Angel and Randy, This is a very nice point! This will be the new point QA-P38. I think the name Cyclocevian Center like Randy proposed it, is a nice
Message 2 of 12 , Jul 1, 2012
Dear Angel and Randy,

This is a very nice point!
This will be the new point QA-P38.
I think the name "Cyclocevian Center" like Randy proposed it, is a nice one.
I has the same structure as QA-P5. Only then "isotomic conjugate" replaced by "cyclocevian conjugate".
I searched for relations with other Quadrangle points.
There hardly are relations but one.
Let S be the intersection point QA-P1.QA-P6 ^ QA-P22.QA-P29.
This point S is also the Involution Center of the line QA-P1.QA-P6 wrt the Reference Quadrangle.
Now QA-P38 lies on the line through S and QA-P11.
This new line QA-P11.QA-P38.S // QA-P1.QA-P32 // QA-P2.QA-P4 // QA-P7.QA-P8 // QA-P12.QA-P24.

About the Centroid of the Cyclocevian Quadrangle:
This point has very complicated coordinates (formulas 51st degree!, when I am not wrong). In a quick scan I could not find linear relations with lines connecting QA-points or relations with other known QA-curves.

I hope for more new points!

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Randy Hutson <rhutson2@...> wrote:
>
> Dear Angel and Chris,
>
> Another construction for this: Let Qi be the cyclocevian conjugate of Pi wrt PjPkPl.Â  QiQjQkQl could be called the 'Cyclocevian Quadrangle' of PiPjPkPl, and its perspector is this new center, which might be called the 'Cyclocevian Center'.
>
> A related center would be the centroid of the Cyclocevian Quadrangle.Â  Coordinates?
>
> Best regards,
>
> Randy Hutson
>
>
>
>
> >________________________________
> > From: Angel <amontes1949@...>
> >To: Hyacinthos@yahoogroups.com
> >Sent: Friday, June 29, 2012 12:04 PM
> >Subject: [EMHL] Re: new centers/items in complete quadrangle and complete quadrilateral
> >
> >
> >Â
> >Dear Chris van Tienhoven,
> >
> >I have a Quadrangle Center that is not currently in EQF.
> >
> >Let P1, P2, P3, P4 be the defining Quadrangle Points.
> >Let S1 = P1.P4 /\ P2.P3, S2 = P1.P3 /\ P2.P4 and S3 = P1.P2/\ P3.P4.
> >Now S1 S2 S3 is the QA-Diagonal Triangle of the Reference Quadrangle.
> >
> >For each vertex Pi, we take the triangle TjTkTl, where Tj the
> >intersection of the sidelinea PkPl with circumcircle of the triangle S1S2S3 (other than S1, S2, S3)
> >Qi = Perspector of the triangle PjPkPl and TjTkTl
> >
> >The "unknown" Quadrangle Center is the common intersection point of lines Pi.Qi
> >
> >1st CT-Coordinate:
> >
> >p(q+r)(p+2q+r)^2(p+q+2r)^2
> >((1/(p+2q+r)^2)((p+r)(-c^4(p+r)^2(q+r)^2+
> >(p+q)^2(a^4(p+r)^2+b^4(q+r)^2))
> >((q+r)^2(2p+q+r)^2SA+(p+r)^2(p+2q+r)^2SB+
> >(p-q)^2(p+q)^2SC))+
> >(1/(p+q+2r)^2)((p+q)(c^4(p+r)^2(q+r)^2+(p+q)^2(a^4(p+r)^2-
> >b^4(q+r)^2))((q+r)^2(2p+q+r)^2SA+
> >(p-r)^2(p+r)^2SB+(p+q)^2(p+q+2r)^2SC)))
> >
> >1st DT-Coordinate: a^2(c^4p^2q^2 + (b^4p^2 - a^4q^2)r^2)
> >
> >Construction GeoGebra:
> >
> >http://amontes.webs.ull.es/geogebra/EQF_QA_Pn.html
> >
> >If ABC is the diagonal triangle of the Quadrangle with a vertex in triangle center X(n), Then the Quadriangle Center obtained here is the TCC-perspector of X(n). See the note just before X(1601) in ETC.
> >
> >Best regards,
> >
> >Angel Montesdeoca
> >
> >--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@> wrote:
> >>
> >> Dear Friends,
> >>
> >> I got some very nice remarks following my announcement of the Encyclopedia of Quadri-Figures.
> >> Seiichi Kirikami attended me on a special property of a Complete Quadrangle.
> >> The Orthopoles of a line with respect to the four component triangles of any complete quadrangle lie on a straight line known as the Orthopolar Line of for the given complete quadrangle (see also "Orthopolar line" in Mathworld).
> >> Definition Orthopole of a random line wrt some Triangle:
> >> If perpendiculars are dropped on any line from the vertices of a triangle, then the perpendiculars to the opposite sides from their perpendicular feet are concurrent at a point called the Orthopole (see also "Orthopole" in Mathworld).
> >>
> >> I tried some things with the Orthopolar Line and found these results:
> >> 1. The 4 Orthopoles of a line wrt the Component Triangles in a Complete Quadrangle are collinear. Let's name this the QA-Orthopole Line.
> >> 2. The 4 Orthopoles of a line wrt the Component Triangles in a Complete Quadrilateral are also collinear. Let's name this the QL-Orthopole Line.
> >> 3. Every QA-Orthopole Line of a line through QA-P4 (Isogonal Center) is a line through QA-P2 (Euler-Poncelet Point).
> >> It looks like this is the only case that a pencil of lines through a point is being transformed into another pencil through a point.
> >> The locus of their mutual crosspoint is a hyperbola through QA-P2 and QA-P4.
> >> Its conic center is the midpoint of QA-P2 and QA-P4.
> >> Both lines are parallel when they are parallel to the asymptotes of this hyperbola.
> >> 4. A QL-Orthopole Line is always a line perpendicular to the Newton Line (QL-L1). Most Special.
> >>
> >> I refer here to some points as coded in the Encyclopedia of Quadri-Figures:
> >> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
> >>
> >> Are there any (synthetic) proofs for 1., 2., 3. and 4. ?
> >> Are there any more related properties known?
> >>
> >> Best regards,
> >>
> >> Chris van Tienhoven
> >>
> >>
> >> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@> wrote:
> >>
> >> > You can find this Encyclopedia of Quadri-Figures (EQF) at:
> >> > http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
> >> >
> >> > The results also can be downloaded in PDF-format at:
> >> >
> >>
> >
> >
> >
> >
> >
>
> [Non-text portions of this message have been removed]
>
• Dear Chris van Tienhoven, In the complete quadriangle determined by the four points P1, P2, P3, P4, the six lines which join them, cut the circumcircle of the
Message 3 of 12 , Jul 2, 2012
Dear Chris van Tienhoven,

In the complete quadriangle determined by the four points P1, P2, P3, P4, the six lines which join them, cut the circumcircle of the diagonal triangle, S1S2S3, into six points (other than S1, S2, S3). The lines joining pairs of these points, which are on opposite sides of the complete quadriangle, are concurrent in the new point "QA-P38", "Cyclocevian Center" like Randy proposed.

Best regards,

Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Angel and Randy,
>
> This is a very nice point!
> This will be the new point QA-P38.
> I think the name "Cyclocevian Center" like Randy proposed it, is a nice one.
> I has the same structure as QA-P5. Only then "isotomic conjugate" replaced by "cyclocevian conjugate".
> I searched for relations with other Quadrangle points.
> There hardly are relations but one.
> Let S be the intersection point QA-P1.QA-P6 ^ QA-P22.QA-P29.
> This point S is also the Involution Center of the line QA-P1.QA-P6 wrt the Reference Quadrangle.
> Now QA-P38 lies on the line through S and QA-P11.
> This new line QA-P11.QA-P38.S // QA-P1.QA-P32 // QA-P2.QA-P4 // QA-P7.QA-P8 // QA-P12.QA-P24.
>
> About the Centroid of the Cyclocevian Quadrangle:
> This point has very complicated coordinates (formulas 51st degree!, when I am not wrong). In a quick scan I could not find linear relations with lines connecting QA-points or relations with other known QA-curves.
>
> I hope for more new points!
>
> Best regards,
>
> Chris van Tienhoven
>
>
> --- In Hyacinthos@yahoogroups.com, Randy Hutson <rhutson2@> wrote:
> >
> > Dear Angel and Chris,
> >
> > Another construction for this: Let Qi be the cyclocevian conjugate of Pi wrt PjPkPl.Â  QiQjQkQl could be called the 'Cyclocevian Quadrangle' of PiPjPkPl, and its perspector is this new center, which might be called the 'Cyclocevian Center'.
> >
> > A related center would be the centroid of the Cyclocevian Quadrangle.Â  Coordinates?
> >
> > Best regards,
> >
> > Randy Hutson
> >
> >
> >
> >
> > >________________________________
> > > From: Angel <amontes1949@>
> > >To: Hyacinthos@yahoogroups.com
> > >Sent: Friday, June 29, 2012 12:04 PM
> > >Subject: [EMHL] Re: new centers/items in complete quadrangle and complete quadrilateral
> > >
> > >
> > >Â
> > >Dear Chris van Tienhoven,
> > >
> > >I have a Quadrangle Center that is not currently in EQF.
> > >
> > >Let P1, P2, P3, P4 be the defining Quadrangle Points.
> > >Let S1 = P1.P4 /\ P2.P3, S2 = P1.P3 /\ P2.P4 and S3 = P1.P2/\ P3.P4.
> > >Now S1 S2 S3 is the QA-Diagonal Triangle of the Reference Quadrangle.
> > >
> > >For each vertex Pi, we take the triangle TjTkTl, where Tj the
> > >intersection of the sidelinea PkPl with circumcircle of the triangle S1S2S3 (other than S1, S2, S3)
> > >Qi = Perspector of the triangle PjPkPl and TjTkTl
> > >
> > >The "unknown" Quadrangle Center is the common intersection point of lines Pi.Qi
> > >
> > >1st CT-Coordinate:
> > >
> > >p(q+r)(p+2q+r)^2(p+q+2r)^2
> > >((1/(p+2q+r)^2)((p+r)(-c^4(p+r)^2(q+r)^2+
> > >(p+q)^2(a^4(p+r)^2+b^4(q+r)^2))
> > >((q+r)^2(2p+q+r)^2SA+(p+r)^2(p+2q+r)^2SB+
> > >(p-q)^2(p+q)^2SC))+
> > >(1/(p+q+2r)^2)((p+q)(c^4(p+r)^2(q+r)^2+(p+q)^2(a^4(p+r)^2-
> > >b^4(q+r)^2))((q+r)^2(2p+q+r)^2SA+
> > >(p-r)^2(p+r)^2SB+(p+q)^2(p+q+2r)^2SC)))
> > >
> > >1st DT-Coordinate: a^2(c^4p^2q^2 + (b^4p^2 - a^4q^2)r^2)
> > >
> > >Construction GeoGebra:
> > >
> > >http://amontes.webs.ull.es/geogebra/EQF_QA_Pn.html
> > >
> > >If ABC is the diagonal triangle of the Quadrangle with a vertex in triangle center X(n), Then the Quadriangle Center obtained here is the TCC-perspector of X(n). See the note just before X(1601) in ETC.
> > >
> > >Best regards,
> > >
> > >Angel Montesdeoca
• Dear Friends, I added the new point QA-P38 to the Encyclopedia of Quadri-Figures (EQF) with the name: Montesdeoca-Hutson Point. See:
Message 4 of 12 , Jul 5, 2012
Dear Friends,

I added the new point QA-P38 to the Encyclopedia of Quadri-Figures (EQF) with the name: Montesdeoca-Hutson Point.

I noticed there is interest in how to find/construct a new Quadrangle/Quadrilateral point.
Here are some tips and tricks:

1. Discern at what level you work:
 Quadrangle (4 points without restriction),
 Quadrilateral (4 lines without restriction) or
 Quadrigon (4 points and 4 cyclically connected lines).

2. When you work with a Quadrangle or Quadrilateral there are 2 very useful sets of Component types:
a. Set of 4 Component Triangles
 P1.P2.P3, P2.P3.P4, P3.P4.P1, P4.P1.P2 in a Quadrangle
 L1.L2.L3, L2.L3.L4, L3.L4.L1, L4.L1.L2 in a Quadrilateral
You can take all triangle properties you are familiar with to combine them 4 times in a Quadrangle or Quadrilateral and look for regularities.

b. Set of 3 Component Quadrigons
 P1.P2.P3.P4, P1.P2.P4.P3, P1.P3.P2.P4 in a Quadrangle
 L1.L2.L3.L4, L1.L2.L4.L3, L1.L3.L2.L4 in a Quadrilateral.
There are very few persons who ever worked with this Component Type, while this is also very useful.
In a Quadrangle you can take all Quadrigon properties as well as Quadrilateral properties to combine them 3 times (for each Component) and look for regularities.
In a Quadrilateral you can take all Quadrigon properties as well as Quadrangle properties to combine them 3 times (for each Component) and look for regularities.

QA-P38 is a nice example of method 2a.
QA-P9 (Miquel Center) is a nice example of method 2b.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
>
> Dear Chris van Tienhoven,
>
> In the complete quadriangle determined by the four points P1, P2, P3, P4, the six lines which join them, cut the circumcircle of the diagonal triangle, S1S2S3, into six points (other than S1, S2, S3). The lines joining pairs of these points, which are on opposite sides of the complete quadriangle, are concurrent in the new point "QA-P38", "Cyclocevian Center" like Randy proposed.
>
> Best regards,
>
> Angel Montesdeoca
• Dear Hyacinthists, The tangents to each of the Pair of Circumscribed Parabolas (QA-2Co1) at the vertices P1, P2, P3, P4 of a complete quadrangle form a pair of
Message 5 of 12 , Jul 5, 2012
Dear Hyacinthists,

The tangents to each of the Pair of Circumscribed Parabolas (QA-2Co1) at the vertices P1, P2, P3, P4 of a complete quadrangle form a pair of complete quadrilaterals.  Question: Are these 2 quadrilaterals in perspective to one another (i.e., do the lines through their corresponding crosspoints concur?).  If so, what is  the perspector?

Best regards,
Randy

>________________________________
> From: Chris Van Tienhoven <van10hoven@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Friday, June 15, 2012 3:31 PM
>Subject: [EMHL] new centers/items in complete quadrangle and complete quadrilateral
>
>

>Dear Friends,
>
>I noticed there are all kind of centers, lines, conics, cubics, etc. in a Complete Quadrilateral/Complete Quadrangle.
>So I started writing a paper on this subject.
>Because I discovered so many new items I decided to make a catalogue of them.
>More than 150 items are catalogued now.
>I gave them all a unique code and name.
>* Jean-Louis' Euler-Poncelet Point (Hyacinthos message 19258) is coded with QA-P2.
>* Jean-Pierre's Homothetic Center (Hyacinthos message 19635) is coded with QA-P4.
>* Apart from the well-known quadrangle centroid QA-P1 there is also a quadrilateral centroid QL-P12.
>I placed them all at my site with references where I found them.
>I also found many new items. These you will find without reference.
>When there are more references please let me know.
>
>You can find this Encyclopedia of Quadri-Figures (EQF) at:
>http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
>The results also can be downloaded in PDF-format at:
>
>Special thanks for Eckart Schmidt who painstakingly checked all items and helped me to eliminate several typos and mistakes and gave me lots of good advice. We had a very nice conversation alternately in German and English.
>
>
>Best regards,
>Chris van Tienhoven
>
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Hyacinthists, The QA-Perimeter Centroid (or some such name) is the centroid of the six segments making up a complete quadrangle.  It can be constructed
Message 6 of 12 , Jul 5, 2012
Dear Hyacinthists,

The 'QA-Perimeter Centroid' (or some such name) is the centroid of the six segments making up a complete quadrangle.  It can be constructed by taking the centroid of the Spieker centers of the four component triangles.

Similarly, the 'QL-Perimeter Centroid' is the centroid of the four segments making up a complete quadrilateral.  It can also be constructed by taking the centroid of the Spieker centers of the four component triangles.

What are the relationships, if any, to existing QA- and QL-points?

Coordinates?

Best regards,
Randy

>________________________________
> From: Chris Van Tienhoven <van10hoven@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Friday, June 15, 2012 3:31 PM
>Subject: [EMHL] new centers/items in complete quadrangle and complete quadrilateral
>
>

>Dear Friends,
>
>I noticed there are all kind of centers, lines, conics, cubics, etc. in a Complete Quadrilateral/Complete Quadrangle.
>So I started writing a paper on this subject.
>Because I discovered so many new items I decided to make a catalogue of them.
>More than 150 items are catalogued now.
>I gave them all a unique code and name.
>* Jean-Louis' Euler-Poncelet Point (Hyacinthos message 19258) is coded with QA-P2.
>* Jean-Pierre's Homothetic Center (Hyacinthos message 19635) is coded with QA-P4.
>* Apart from the well-known quadrangle centroid QA-P1 there is also a quadrilateral centroid QL-P12.
>I placed them all at my site with references where I found them.
>I also found many new items. These you will find without reference.
>When there are more references please let me know.
>
>You can find this Encyclopedia of Quadri-Figures (EQF) at:
>http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
>The results also can be downloaded in PDF-format at:
>
>Special thanks for Eckart Schmidt who painstakingly checked all items and helped me to eliminate several typos and mistakes and gave me lots of good advice. We had a very nice conversation alternately in German and English.
>
>
>Best regards,
>Chris van Tienhoven
>
>
>
>
>

[Non-text portions of this message have been removed]
• ... That s a hybrid word, combining the Latin prefix quadri with the Greek stem gon . The purely Greek version is tetragon . According to Heath s
Message 7 of 12 , Jul 8, 2012
On 5-7-2012, at 11.07 PM, Chris Van Tienhoven wrote:

> ….
> • Quadrangle (4 points without restriction),
> • Quadrilateral (4 lines without restriction) or
> • Quadrigon (4 points and 4 cyclically connected lines)….

That's a hybrid word, combining the Latin prefix "quadri" with the Greek stem "gon". The purely Greek version is "tetragon". According to Heath's Euclid, vol.1, p.188 (note on I Def.22), the Greek word "tetragonon" had come to mean "square" in Euclid's time, but there is some evidence of its earlier use for any four-sided figure.

FWIW in my teaching I use the wortd "tetragon" for the Euclidean figure, reserving Steiner's names "quadrangle" and "quadrilateral" for use in projective geometry.

Ken Pledger.
• Dear Friends, I added a new property wrt a conic in a complete quadrangle in EQF. See:
Message 8 of 12 , Jul 21, 2012
Dear Friends,

I added a new property wrt a conic in a complete quadrangle in EQF.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Friends,
>
> I noticed there are all kind of centers, lines, conics, cubics, etc. in a Complete Quadrilateral/Complete Quadrangle.
> So I started writing a paper on this subject.
> Because I discovered so many new items I decided to make a catalogue of them.
> More than 150 items are catalogued now.
> I gave them all a unique code and name.
> * Jean-Louis' Euler-Poncelet Point (Hyacinthos message 19258) is coded with QA-P2.
> * Jean-Pierre's Homothetic Center (Hyacinthos message 19635) is coded with QA-P4.
> * Apart from the well-known quadrangle centroid QA-P1 there is also a quadrilateral centroid QL-P12.
> I placed them all at my site with references where I found them.
> I also found many new items. These you will find without reference.
> When there are more references please let me know.
>
> You can find this Encyclopedia of Quadri-Figures (EQF) at:
> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
> The results also can be downloaded in PDF-format at:
>
> Special thanks for Eckart Schmidt who painstakingly checked all items and helped me to eliminate several typos and mistakes and gave me lots of good advice. We had a very nice conversation alternately in German and English.
>
>
> Best regards,
> Chris van Tienhoven
>
• I repeat my previous message because of typo in hyperlink. Dear Friends, I added a new property wrt a conic in a complete quadrangle in EQF. See:
Message 9 of 12 , Jul 21, 2012
I repeat my previous message because of typo in hyperlink.

Dear Friends,

I added a new property wrt a conic in a complete quadrangle in EQF.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Friends,
>
> I noticed there are all kind of centers, lines, conics, cubics, etc. in a Complete Quadrilateral/Complete Quadrangle.
> So I started writing a paper on this subject.
> Because I discovered so many new items I decided to make a catalogue of them.
> More than 150 items are catalogued now.
> I gave them all a unique code and name.
> * Jean-Louis' Euler-Poncelet Point (Hyacinthos message 19258) is coded with QA-P2.
> * Jean-Pierre's Homothetic Center (Hyacinthos message 19635) is coded with QA-P4.
> * Apart from the well-known quadrangle centroid QA-P1 there is also a quadrilateral centroid QL-P12.
> I placed them all at my site with references where I found them.
> I also found many new items. These you will find without reference.
> When there are more references please let me know.
>
> You can find this Encyclopedia of Quadri-Figures (EQF) at:
> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
> The results also can be downloaded in PDF-format at: