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Re: Cyclic quadrilateral, correction

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  • Aad Goddijn
    Dear Paul, Alex, 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...
    Message 1 of 1 , Jan 5, 2003
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      Dear Paul, Alex,

      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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