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[EMHL] Re: Just a center - as always

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  • Hauke Reddmann
    ... drawing by ... Lineal.) ... THX for the tip. DLed. Can someone check the following? It doesn t work with actual test values, and I must have made a stupid
    Message 1 of 11 , Dec 1, 2008
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      >
      > Consider also to invest into a geometry software instead of
      drawing "by
      > eye".
      > (there are many free software, one of the best is IMHO Zirkel und
      Lineal.)
      >
      THX for the tip. DLed.

      Can someone check the following? It doesn't work with
      actual test values, and I must have made a stupid error
      somewhere.

      Call the triangle abc. Draw isosceles triangles
      buc,cva and awb and call cot(angle(buc)/2)=u for short.
      Assume AU,BV,CW meet in O, make this the origin and
      use vector algebra. (A is the vector to a etc.)

      Then U=A*j, and since (U-(B+C)/2)(B-C)=0, 2j=(BB-CC)/(AB-AC).

      ||U-(B+C)/2||/||(B-C)/2||=cos(buc/2)/sin(buc/2)=u.

      Thus (2j)^2*AA+BB+CC+2*BC-2*(2j)*(AB+AC)=a^2*u.
      (plus 2 "rotated" equations" with v,w)
      Also (A-B)^2=AA+BB-2*AB=c^2 (+2 rotated)
      Finally (AxB).C=0. Eliminate the scalar products to get
      the master equation in u,v,w.
      Which is a complete mess, thus it's better to use the
      u,v,w you want to test and NSolve[] the mess instead :-)
      (So far checking with the circumcenter, where the angles
      are pi-@,... failed. I try again.)

      Hauke
    • Hauke Reddmann
      ... UNCLE OF EDIT: Flunked a factor of 2. Now it works with the circumcenter (pi-@,pi-ß,pi-y). (@,ß,y) does NOT work. Double drat. (2@,2ß,2y) as well as
      Message 2 of 11 , Dec 1, 2008
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        > (So far checking with the circumcenter, where the angles
        > are pi-@,... failed. I try again.)
        >
        UNCLE OF EDIT:
        Flunked a factor of 2. Now it works with the circumcenter
        (pi-@,pi-ß,pi-y). (@,ß,y) does NOT work. Double drat.
        (2@,2ß,2y) as well as (pi-2@,pi-2ß,pi-2y) and any (x,x,x)
        seem to work.

        Hauke
      • Philippe
        Hi Hyacinthians, ... [ That is draw on ABC the isosceles triangles BUC, CVA, AWB, with angles U,VW equal to x (similar isosceles triangles) Then AU, BV, CW
        Message 3 of 11 , Dec 1, 2008
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          Hi Hyacinthians,

          Hauke Reddmann wrote:
          >> ... and any (x,x,x) seem to work.
          >

          [ That is draw on ABC the isosceles triangles BUC, CVA, AWB,
          with angles U,VW equal to x (similar isosceles triangles)
          Then AU, BV, CW concur in some point X ]

          Yes. And then the locus of X is Kiepert Hyperbola, by definition.

          For instance.
          http://faculty.evansville.edu/ck6/bstud/kiepert.html
          http://mathworld.wolfram.com/KiepertHyperbola.html

          etc.

          There are many known points on this hyperbola.
          Have fun with finding your own point ;-)

          Regards.
          Philippe.
        • Hauke Reddmann
          ... THX, I would have asked that eventually anyway (my memory - x,x,x works - *didn t* fail me this time :-) This leaves (for ending this thread :-) the
          Message 4 of 11 , Dec 4, 2008
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            --- In Hyacinthos@yahoogroups.com, "Philippe" <chephip@...> wrote:
            >
            > Yes. And then the locus of X is Kiepert Hyperbola, by definition.
            >
            > For instance.
            > http://faculty.evansville.edu/ck6/bstud/kiepert.html
            > http://mathworld.wolfram.com/KiepertHyperbola.html
            >
            THX, I would have asked that eventually anyway
            (my memory - x,x,x works - *didn't* fail me this time :-)

            This leaves (for ending this thread :-) the question
            which are the points corresponding to isosceles
            triangles with peak angles 2@,2ß,2y resp. pi-2@...
            Would do it myself but my PC is currently in repair.
            I expect some classic points.

            Hauke
          • Nikolaos Dergiades
            Dear Hauke, If I understood your problem we have the following: Let 2p, 2q, 2r be the peak angles of the isosceles triangles A BC, B CA, C AB outside ABC and x
            Message 5 of 11 , Dec 5, 2008
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              Dear Hauke,
              If I understood your problem we have the following:
              Let 2p, 2q, 2r be the peak angles of the isosceles
              triangles A'BC, B'CA, C'AB outside ABC and
              x = cot(p), y = cot(q), z = cot(r)
              X = cot(A), Y = cot(B), Z = cot(C)
              If AA' meets BC at A" then
              BA"/A"C = (ABA')/(AA'C)
              = AB.BA'.sin(B+90-p)/AC.CA'.sin(C+90-p)
              =sinC.cos(p-B)/sinB.cos(p-C)
              =(cotp.cotB+1)/(cotp.cotC+1)
              =(xY+1)/(xZ+1)
              From Ceva's theorem AA", BB", CC" are concurrent
              iff [(x.Y+1)/(x.Z+1)].[(y.Z+1)/(y.X+1)].[(zX+1)/(zY+1)]=1

              An obvious solution is x = y = z or p = q = r = s
              and the perspector P is on the Kiepert hyperbola.
              s = pi/2 then P = X(2)
              s = 0 then P = X(4)
              s = pi/6 then P = X(13)
              s = -pi/6 then P = X(14)
              s = pi/3 then P = X(17)
              s = -pi/3 then P = X(18)
              s = pi/4 then P = X(485)
              s = -pi/4 then P = X(486)

              Another partial solution is
              p = arccot(cot(A)/n)
              q = arccot(cot(B)/n)
              r = arccot(cot(C)/n)
              where arccot is the inverse function of cot
              If n = 1 then p = A, q = B, r = C and P = X(5)
              If n = -1 then p = -A, q = -B, r = -C and P = X(3).

              Another partial solution is p = (pi-A)/2, q = (pi-B)/2, r = (pi-C)/2
              and P = X(1)

              Best regards
              Nikos Dergiades



              > >
              > > Yes. And then the locus of X is Kiepert Hyperbola, by
              > definition.
              > >
              > > For instance.
              > > http://faculty.evansville.edu/ck6/bstud/kiepert.html
              > > http://mathworld.wolfram.com/KiepertHyperbola.html
              > >
              > THX, I would have asked that eventually anyway
              > (my memory - x,x,x works - *didn't* fail me this time
              > :-)
              >
              > This leaves (for ending this thread :-) the question
              > which are the points corresponding to isosceles
              > triangles with peak angles 2@,2ß,2y resp. pi-2@...
              > Would do it myself but my PC is currently in repair.
              > I expect some classic points.
              >
              > Hauke
              >




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            • Nikolaos Dergiades
              Dear Alexey, thank you very much. You sent your message only to me and not to hyacinthos. So I tried to answer privately without success. My message returned.
              Message 6 of 11 , Dec 6, 2008
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                Dear Alexey,
                thank you very much.
                You sent your message only to me
                and not to hyacinthos. So I tried to answer
                privately without success. My message returned.
                You are right.
                I restate your property as following:
                If P, Q, R are three points on Kiepert hyperbola
                and p, q, r are the corresponding base angles of
                the isosceles triangles giving these points
                and x = cot(p), y = cot(q), z = cot(r)
                then in barycentrics we have
                P = (1/(xS+S_A) : 1/(xS+S_B) : (1/(xS+S_C)
                Q = (1/(yS+S_A) : 1/(yS+S_B) : (1/(yS+S_C)
                R = (1/(zS+S_A) : 1/(zS+S_B) : (1/(zS+S_C)
                The isogonal conjugate of any vertex of triangle PQR
                lies on the opposite side
                (e.g. the isogonal conjugate P' of P lies on line QR)
                iff xy + yz + zx = 1
                or x = (1-yz)/(y+z) or cot(p) = cot(-q-r)
                or p + q + r = k.pi where we can have k = 0, +-1, +-2
                For example if
                p = pi/2 then P = X(4) and P' = X(3)
                q = pi/3 then Q = X(13) and Q' = X(15)
                r = pi/6 then R = X(17) and R' = X(61)
                since pi/2 + pi/3 + pi/6 = 1.pi
                then by the previous property
                X(3) lies on the line X(13)X(17)
                X(15) lies on the line X(4)X(17)
                X(61) lies on the line X(4)X(13)

                Best regards
                Nikos Dergiades


                > Dear colleagues!
                > I proved interesting property of Kiepert hyperbola.
                > Let $ABC'$, $BCA'$, $CAB'$ be similar isosceles
                > triangles with $
                > \angle C'AB=\phi$. We regard $\phi>0$
                > when these triangles are
                > outside $ABC$ and $\phi<0$ in the opposite case, so
                > $-\pi/2\leq\phi
                > \leq\pi/2$. Note as $X(\phi)$,
                > $X'(\phi)$ the perspective center of
                > $ABC$ and $A'B'C'$ and the point of line $OL$
                > isogonally conjugated
                > to it. Then
                > $X(\alpha)$, $X(\beta)$, $X'(\gamma)$ are
                > collinear iff $\alpha+\beta+
                > \gamma=0$.
                > Another interesting property. Triangles $ABC$ and
                > $A'B'C'$ are
                > orthologic. One of orthology centers is $O$, and stcond is
                > $X(\pi/2-
                > \phi)$, so it also lies on Kiepert hyperdola.
                >
                > Sincerely Alexey





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              • Hauke Reddmann
                ... Yup, I also dimly remembered there was a classic solution with half-angles. Your arccot formula explains many things, although the cases 2A,2B,2C and
                Message 7 of 11 , Dec 7, 2008
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                  --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades
                  <ndergiades@...> wrote:
                  >
                  > Another partial solution is
                  > p = arccot(cot(A)/n)
                  > q = arccot(cot(B)/n)
                  > r = arccot(cot(C)/n)
                  > where arccot is the inverse function of cot
                  > If n = 1 then p = A, q = B, r = C and P = X(5)
                  > If n = -1 then p = -A, q = -B, r = -C and P = X(3).
                  >
                  > Another partial solution is p = (pi-A)/2, q = (pi-B)/2, r = (pi-C)/2
                  > and P = X(1)
                  >
                  Yup, I also dimly remembered there was a classic solution
                  with half-angles. Your arccot formula explains many things,
                  although the cases 2A,2B,2C and -2A,-2B,-2C (modulo 2pi)
                  are still not covered by it.
                  ("Just" computing the coordinates of O for the Kimberling
                  triangle and looking it up smells like problems - the
                  double of an obtuse angle... :-)

                  Hauke
                • chris.vantienhoven
                  Dear Hauke, Dear Nikolaos, In addition to the solutions (and notation) of Nikolaos here are some extra solutions though not based on isosceles triangles but on
                  Message 8 of 11 , Dec 8, 2008
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                    Dear Hauke, Dear Nikolaos,

                    In addition to the solutions (and notation) of Nikolaos here are some
                    extra solutions though not based on isosceles triangles but on
                    similar triangles.

                    if s = D/2 then P = X(1)
                    if s = D then P = X(6)
                    (D = opposite angle of side on triangle)

                    if s = E then P = X(4)
                    if s = 2pi/3 - E then P = X(15)
                    if s = -2pi/3 + E then P = X(16)
                    if s = pi - 2E or pi/2 - E then P = X(3)
                    (E = adjacent angle of side on triangle)

                    Further, in general can be stated that:
                    if s = F then P = Q
                    (F = corresponding angle in pedal triagle of isogonal conjugate of Q)
                    This means that there is a solution for any point (just a center) by
                    taking the isogonal conjugate, then its pedal triangle, then by
                    rising the angles of the pedal triangle on relevant sides.

                    I hope this is helpfull.
                    Regards,

                    Chris van Tienhoven



                    --- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades
                    <ndergiades@...> wrote:
                    >
                    > Dear Hauke,
                    > If I understood your problem we have the following:
                    > Let 2p, 2q, 2r be the peak angles of the isosceles
                    > triangles A'BC, B'CA, C'AB outside ABC and
                    > x = cot(p), y = cot(q), z = cot(r)
                    > X = cot(A), Y = cot(B), Z = cot(C)
                    > If AA' meets BC at A" then
                    > BA"/A"C = (ABA')/(AA'C)
                    > = AB.BA'.sin(B+90-p)/AC.CA'.sin(C+90-p)
                    > =sinC.cos(p-B)/sinB.cos(p-C)
                    > =(cotp.cotB+1)/(cotp.cotC+1)
                    > =(xY+1)/(xZ+1)
                    > From Ceva's theorem AA", BB", CC" are concurrent
                    > iff [(x.Y+1)/(x.Z+1)].[(y.Z+1)/(y.X+1)].[(zX+1)/(zY+1)]=1
                    >
                    > An obvious solution is x = y = z or p = q = r = s
                    > and the perspector P is on the Kiepert hyperbola.
                    > s = pi/2 then P = X(2)
                    > s = 0 then P = X(4)
                    > s = pi/6 then P = X(13)
                    > s = -pi/6 then P = X(14)
                    > s = pi/3 then P = X(17)
                    > s = -pi/3 then P = X(18)
                    > s = pi/4 then P = X(485)
                    > s = -pi/4 then P = X(486)
                    >
                    > Another partial solution is
                    > p = arccot(cot(A)/n)
                    > q = arccot(cot(B)/n)
                    > r = arccot(cot(C)/n)
                    > where arccot is the inverse function of cot
                    > If n = 1 then p = A, q = B, r = C and P = X(5)
                    > If n = -1 then p = -A, q = -B, r = -C and P = X(3).
                    >
                    > Another partial solution is p = (pi-A)/2, q = (pi-B)/2, r = (pi-C)/2
                    > and P = X(1)
                    >
                    > Best regards
                    > Nikos Dergiades
                    >
                    >
                    >
                    > > >
                    > > > Yes. And then the locus of X is Kiepert Hyperbola, by
                    > > definition.
                    > > >
                    > > > For instance.
                    > > > http://faculty.evansville.edu/ck6/bstud/kiepert.html
                    > > > http://mathworld.wolfram.com/KiepertHyperbola.html
                    > > >
                    > > THX, I would have asked that eventually anyway
                    > > (my memory - x,x,x works - *didn't* fail me this time
                    > > :-)
                    > >
                    > > This leaves (for ending this thread :-) the question
                    > > which are the points corresponding to isosceles
                    > > triangles with peak angles 2@,2ß,2y resp. pi-2@
                    > > Would do it myself but my PC is currently in repair.
                    > > I expect some classic points.
                    > >
                    > > Hauke
                    > >
                    >
                    >
                    >
                    >
                    > ___________________________________________________________
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