There was a mistake in my last reply, as you probably noticed.

I combined the right homotheties, but to quick and not in the right way...

Combining the homotheties (O, 3) and (G,-1/3) gives a homothety (H, -1),

with OH = 2 OG; because this point H is invariant after applying (G,-1/3)

and then (O, 3).

Combining (O, 3/2) with (G, -1/3) gives (N, -2) with ON = 4/3 * OG. ,

This confirms your results.

The O-G-N-H propotions are not the same as on the normal Eulerline and my

last remark is

to be forgotten as soon as possible!

Greetings,

Aad

----- Original Message -----

From: <Hyacinthos@yahoogroups.com>

To: <Hyacinthos@yahoogroups.com>

Sent: Saturday, January 04, 2003 5:09 PM

Subject: [EMHL] Digest Number 936

>

> There are 7 messages in this issue.

>

> Topics in this digest:

>

> 1. Re: Cyclic quadrilateral

> From: "Paul Yiu <yiu@...>" <yiu@...>

> 2. Re: Re: Cyclic quadrilateral

> From: Alexander Bogomolny <alexb@...>

> 3. Re: Nine-point circles concurring at Feuerbach point (was: Cyclic

quadrilateral)

> From: Alexander Bogomolny <alexb@...>

> 4. Re: Some theorems on Miquel points

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> 5. Re: Some theorems on Miquel points (typo)

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> 6. Re: Some theorems on Miquel points (new typo)

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> 7. Five concyclic points

> From: "Ercole Suppa" <ercsuppa@...>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 1

> Date: Fri, 03 Jan 2003 20:34:44 -0000

> From: "Paul Yiu <yiu@...>" <yiu@...>

> Subject: Re: Cyclic quadrilateral

>

> Dear Alex and friends on Hyacinthos,

>

> Happy New Year.

>

> [AB]:...

> >

> > Further, for a cyclic quadrilateral ABCD, the quadrilateral of

> > the orthocenters, is homothetic to ABCD, with the factor of -1.

> > Obviously, the quadrilateral of centroids is also homothetic to

> > ABCD with the factor of -1/3. It appears that the quadrilateral

> > of the 9-point centers is also homothetic to ABCD with the factor

> > of -1/2.

> >

> > After toying for a while with Java, I conjecture that there's

> > a line that might be rightfully called Euler's line of a cyclic

> > quadrilateral. Let's denote the above centers of homotheties

> > successively H, N, G. There's of course the common circumcenter O.

> > Then the four points H, N, G, O are collinear such that HG = GO

> > and 2*HN = NO.

>

> Dear Alex,

>

> The line in question is the one joining the center of the circle to

> the centroid of the quadrilateral, the intersection of the

> lines joining the midpoints of the two pairs of opposite sides.

>

> It is convenient to use complex number coordinates. We put the

> circumcenter at the origin, and represent the vertices A, B, C, D

> of the quadrilateral by unit complex numbers a, b, c, d.

>

> It is well known that the orthocenter of triangle abc is a+b+c etc.

>

> We consistently denote by P(t) the point P on the Euler line of a

> triangle with

> OP : OH= t : 1. [Thus, P(0) = O, P(1/2) = ninepoint center, P(1/3)

> = centroid etc.]

>

> Thus, P(t) of triangle bcd is t(b+c+d),

> P(t) of triangle cda is t(c+d+a),

> P(t) of triangle dab is t(d+a+b), and

> P(t) of triangle abc is t(a+b+c).

>

> The quadrilateral of P(t)'s of the four triangles bcd, cda, dab, abc

> is homothetic to the quadrilateral abcd with factor -t. The center of

> homothety is the point Q(t) such that P(t)Q(t) : AQ(t) = -t : 1. This

> means that AQ(t) : Q(t)P(t) = 1 : t, and

>

> Q(t) = t(a+b+c+d)/(1+t).

>

> Note that (a+b+c+d)/4 is the centroid G of the quadrilateral, so

> that this center of homothety traverses the line OG as claimed.

>

> =======

>

> Now if we fix a, b, c and let d vary, then the locus of Q(t)=z is a

> circle, since from z = t(a+b+c+d)/(1+t), we have

>

> z - t(a+b+c)/(1+t) = t/(1+t).d.

>

> The center of the circle is P(t/(1+t)) of triangle ABC, and the

> radius is t/(1+t) times the circumradius of ABC.

>

> For example, the center of homothety of the quadrilateral of

> nine-points, with t = 1/2, traverses the circle, center P(1/3) =

> centroid, radius R/3.

>

> Best regards

> Sincerely

> Paul

>

>

>

>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 2

> Date: Fri, 03 Jan 2003 17:14:23 -0500

> From: Alexander Bogomolny <alexb@...>

> Subject: Re: Re: Cyclic quadrilateral

>

> Dear Paul:

>

> Thank you. It could not be clearer.

>

> > The center of the circle is P(t/(1+t)) of triangle ABC, and the

> > radius is t/(1+t) times the circumradius of ABC.

> >

> > For example, the center of homothety of the quadrilateral of

> > nine-points, with t = 1/2, traverses the circle, center P(1/3) =

> > centroid, radius R/3.

>

> If f(t) = t/(1+t), then

>

> f(0) = 0, f(1) = 1/2, f(1/2) = 1/3, and f(1/3) = 1/4. So indeed

>

> f(1) - f(1/3) = f(1/3) - f(0) and

> 2*(f(1) - f(1/2)) = f(1/2) - f(0).

>

> All the best,

> Alex

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 3

> Date: Fri, 03 Jan 2003 17:55:44 -0500

> From: Alexander Bogomolny <alexb@...>

> Subject: Re: Nine-point circles concurring at Feuerbach point (was: Cyclic

quadrilateral)

>

> Dear Darij:

>

> "Darij Grinberg " wrote:

> >

> > >> "If four finite points be given, whereof no three

> > >> are collinear, which are not the vertices and the

> > >> orthocenter of a triangle, the four nine-point

> > >> circles which they determine three by three, and

> > >> the pedal circle of each with regard to the

> > >> triangle of the other three, are concurrent."

> >

> > Somehow this must be connected with the famous Fontene

> > theorems,

>

> Yes, Coolidge mentions Fontene several times in the

> neighborhood of that theorem. For example in a footnote:)

>

> "Fontene speaks of this as a well-known theorem."

>

> From the above Coolidge derives Fontene's theorem about the

> pedal circle of a point on a line through the circumcenter.

> The isogonal conjugate of such a point will trace a conic

> through the vertices and the orthocenter of the triangle.

> So that the conic is bound to be a rectangular hyperbola.

> And then refer to Jean-Pierre's message #6252.

>

> > but I have no time to look further at it. I

> > am sure that it can be proven elementarily

>

> This is what I hope for.

>

> > What I further know on the four pedal triangles is that

> > they are all similar (this can be proven easily by

> > calculation of lengths).

> >

> > Here is a nice theorem I have discovered:

> >

> > Let I be the incenter of ABC. Then, the Feuerbach

> > point of ABC (i. e. the point of tangency of the

> > incircle with the nine-point circle) lies on the

> > nine-point circles of triangles BIC, CIA, AIB.

> >

>

> Both are indeed very nice results. Thank you.

>

> All the best,

> Alex

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 4

> Date: Sat, 04 Jan 2003 09:04:52 -0000

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> Subject: Re: Some theorems on Miquel points

>

> Dear Darij,

> I've tried to find an easy synthetic proof of Braun theorem

> > The Miquel point configuration is defined as follows:

> >

> > On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'

> are

> > chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we

> > know, these circles have a common point P (the Miquel point).

> >

> > Now the two theorems:

> >

> > 2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A',

> CA'B'

> > [it is well-known that triangles MaMbMc and ABC are similar], let

> M'

> > be the circumcenter of triangle MaMbMc, and M the circumcenter of

> > triangle ABC. Then M'M = M'P.

> >

> > [This is a theorem by Peter Baum; proposed in the little German

> > periodical "Die Wurzel" (see the website wurzel.org), in the 12/98

> > issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution

> > was quite involved.]

>

> PaPbPc is the pedal triangle of P.

> A direct similitude with center P, angle T, ratio 1/cos(T) maps

> PaPbPc to A'B'C'.

> As the circumcenter of PPbPc is the midpoint A'' of AP, the

> similitude maps this midpoint to Ma; hence the similitude maps the

> midpoint I of PM (circumcenter of A''B''C'') to M'(circumcenter of

> MaMbMc).

> As PM'/PI = cos(T) and <IPM' = T, it follows that M' lies on the

> perpendicular bisector of PM.

> Friendly. Jean-Pierre

>

>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 5

> Date: Sat, 04 Jan 2003 09:06:42 -0000

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> Subject: Re: Some theorems on Miquel points (typo)

>

> Dear Darij, I wrote

>

> > I've tried to find an easy synthetic proof of Braun theorem

> Of course, I was meaning Baum theorem.

> Friendly. Jean-Pierre

>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 6

> Date: Sat, 04 Jan 2003 09:10:47 -0000

> From: "jpehrmfr <jean-pierre.ehrmann@...>"

<jean-pierre.ehrmann@...>

> Subject: Re: Some theorems on Miquel points (new typo)

>

> We have PM'/PI = 1/cos(T) and not PM'/PI.

> With my apologizes. Jean-Pierre

>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

> Message: 7

> Date: Sat, 4 Jan 2003 17:22:51 +0100

> From: "Ercole Suppa" <ercsuppa@...>

> Subject: Five concyclic points

>

> Dear Hyacinthists,

> can you "synthetically" prove this problem:

>

> PROBLEM: Let P, Q two points respectively on the sides BC and CD of the

> square ABCD

> such that PQ is tangent to circle with center A amd radius AB. The

segments

> AP and AQ intersect the diagonal BD in R and S. Prove that the point P, C,

> Q,

> R and S are concyclic.

>

> I have found only a trigonometric solution!

>

> Happy New Year to all!

> Ercole Suppa

>

>

>

> ________________________________________________________________________

> ________________________________________________________________________

>

>

>

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>

>