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Re: [EMHL] Quadrilateral and triangle

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  • Francois Rideau
    Dear Hauke The bicentric configuration is well known for a long time. We start with 2 circles and with centers O and o, radii R and r and d =
    Message 1 of 14 , Sep 8, 2007
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      Dear Hauke
      The bicentric configuration is well known for a long time.
      We start with 2 circles <Gamma> and <gamma> with centers O and o, radii R
      and r and d = Oo.
      Suppose we have a "quadrilateral" ABCD inscribed in circle <Gamma> and such
      that sides AB, BC, CD, DA are tangent to circle <gamma> respectively in a b,
      c, d.
      Then we have 2 cases:
      1° lines ac and bd are orthogonal, diagonal triangle EFG is autopolar wrt
      circle <Gamma>, so the center O of circle <Gamma> is the orthocenter of
      triangle EFG.
      2° lines ac and bd are parallel, A and C are symmetric wrt Oo and B and D
      are also symmetric wrt Oo

      Of course for any circles <Gamma> and <gamma>, in general there is no such
      quadrilateral ABCD, so there is some condition necessary and sufficient to
      be fulfilled by the 2 circles for the existence of ABCD.
      In the first case , this the Euler condition :
      (d^2 - R^2)^2 = 4 R^2 r^2 or d^2 - R^2 = (+ -)2Rr
      In the second case, the center o of circle <gamma> must be on circle
      <Gamma>.

      In every case where these conditions are fulfilled, then there is infinitely
      many such quadrilaterals.
      As I think that is the first case which is interesting to you, notice that
      in this case, one of the vertices of the diagonal triangle EFG, for example
      E = AC /\ BD, is fixed, the other F and G are somewhere on the polar line of
      E wrt circle <Gamma>.
      I let you find by yourself the properties of the second case.

      I think you can find these configurations in every good books in geometry,
      for example in the Morley book Inversive geometry with elementary proofs.

      Now you want some parametrization for A, B, C, D and this problem is all but
      easy for analytic proofs are based on the theory of elliptic functions. I
      remember to have read a book on elliptic functions with all a chapter
      devoted to the Poncelet porism.
      Friendly
      Francois

      On 9/7/07, Hauke Reddmann <fc3a501@...> wrote:
      >
      > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
      > "Francois Rideau"
      > <francois.rideau@...> wrote:
      > >
      > > Dear Hauke
      > > Thanks for the precision.
      > > I know what is a bicentric quadrilateraln that is to say a
      > quadrilateral
      > > with a circumcircle <Gamma> and an incircle <gamma> with the
      > Poncelet porism
      > > on the top.
      > > So what is your problem?
      >
      > To find out which special points the circle centers are
      > *with respect to the diagonal triangle*. Neither can I
      > prove that the circumcenter of ABCD is the orthocenter
      > of EFG nor I can even guess what the incenter is, again
      > with respect to EFG. Because the algebraic approach is
      > hopelessly convoluted, and how shall I construct a
      > bicentric quadrilateral whose diagonal triangle is
      > the Kimberling triangle 6-9-13 (the usual approach to
      > get a hint of the center)? Does anyone have a parametric
      > formula for the points ABCD? Otherwise my last chance would
      > be an 8-dimensional trial-and-error search. :-O
      >
      > Hauke
      >
      >
      >


      [Non-text portions of this message have been removed]
    • Eisso J. Atzema
      Dear Hauke, you are completely right that for every (obtuse) triangle there is a bicentric quadrilateral which has this triangle for its diagonal points
      Message 2 of 14 , Sep 9, 2007
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        Dear Hauke,

        you are completely right that for every (obtuse) triangle there is a
        bicentric quadrilateral which has this triangle for its diagonal points
        triangle. Here is a geometric construction (assuming that ABCD is
        convex) that might need more explanation, but this is what I have time
        to write up now. Feel free to ask for clarification:

        1) note that for every complete quadrilateral ABCD its diagonal triangle
        EFG (E=AC\capBD, G=AB\cap CD, F=AD\capBC, with the obtuse angle at E)
        is self-polar with respect to every conic circumscribing ABCD. If ABCD
        is cyclic, that means that EFG has to be self-polar w.r.t. a circle. But
        a necessary condition for a triangle to be self-polar w.r.t. a circle is
        that the center of the circle is the orthocenter of the triangle. The
        condition is not sufficient, but the radius of the circle is easily
        determined as the product of distance from the circle center to any
        vertex and that of the center's distance to the opposite side has to be
        equal to the square of the radius.

        2) If ABCD is to be cyclic, then its nine-point conic is an orthogonal
        hyperbola which has its center at the center point O of the quadrilateral.
        a. It is a known theorem that for ABCD cyclic, O lies on the nine-point
        circle of EFG. This means that the locus of all possible center points
        is well-determined.
        b. For any orthogonal hyperbola, conjugation is reflection in its
        asymptotes. Since for every quadrilateral the diameters of its
        nine-point parallel to a pair of opposite sides are conjugate, it
        follows that the directions of the three pairs of opposite sides of a
        cyclic ABCD are obtained by reflection in the asymptotes of the
        nine-point conic. Particularly, this means that the angle bisectors at
        E, F and G are all parallel to the asymptotes of the nine-point conic.

        3) If ABCD is to be inscribable as well, the interior angle bisectors at
        F and G need to meet at the incenter of the quadrilateral. Since these
        two angle bisectors are perpendicular by 2b, it follows that the
        incenter lies on the circle that has the long side FG of EFG for its
        diameter. Moreover, it is known for any bicentric quadrilateral, its
        incenter lies on the line connecting its diagonal point E to its
        circumcenter. This completely determines the position of the incenter of
        any bicentric quadrilateral ABCD which has EFG for its diagonal triangle.

        The construction of ABCD now follows by extending the line connecting O
        with the mid-point E' of FG. One of the points of intersection of this
        line with the nine-point circle of EFG is the point O of the
        quadrilateral ABCD we are looking for (the other point of intersection
        might correspond to an escribable quadrilateral). Since E,F,G are on the
        nine-point conic of any ABCD of which they are the diagonal points,
        these points together with O completely determine the nine-point conic
        of ABCD. The six mid-points of the sides of ABCD are now found by noting
        that any bimedian of ABCD bisects the corresponding side of EFG
        (extended Gauss-Bodenmiller). Since these mid-points are also on the
        nine-point conic, they are the points of intersections of the lines
        connecting O with the mid-points of EFG. Once we have the mid-points of
        ABCD, we find the quadrilateral itself by connecting E, F, and G with
        the corresponding mid-points. This works out nicely without any
        ambiguity and we have our bicentric quadrilateral ABCD.

        The whole construction can be done in SketchPad, provided you have the
        right macros to draw a conic section and to find the points of
        intersection of a conic section with a straight line (both of which are
        part of my standard arsenal of macros). I can post a sketch if you want one.

        Eisso




        Hauke Reddmann wrote:
        >
        >
        > To find out which special points the circle centers are
        > *with respect to the diagonal triangle*. Neither can I
        > prove that the circumcenter of ABCD is the orthocenter
        > of EFG nor I can even guess what the incenter is, again
        > with respect to EFG. Because the algebraic approach is
        > hopelessly convoluted, and how shall I construct a
        > bicentric quadrilateral whose diagonal triangle is
        > the Kimberling triangle 6-9-13 (the usual approach to
        > get a hint of the center)? Does anyone have a parametric
        > formula for the points ABCD? Otherwise my last chance would
        > be an 8-dimensional trial-and-error search. :-O
        >
        > Hauke
        >
        >

        --

        ========================================
        Eisso J. Atzema, Ph.D.
        Department of Mathematics & Statistics
        University of Maine
        Orono, ME 04469
        Tel.: (207) 581-3928 (office)
        (207) 866-3871 (home)
        Fax.: (207) 581-3902
        E-mail: atzema@...
        ========================================
      • Francois Rideau
        Dear Eisso Thank you for your wonderful construction and in any case cheerup for your deciphering Hauke s thought for I have not understood what he was asking
        Message 3 of 14 , Sep 10, 2007
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          Dear Eisso
          Thank you for your wonderful construction and in any case cheerup for your
          deciphering Hauke's thought for I have not understood what he was asking
          for.
          Of course this bicentric configuration is well known as I have said it
          before but I did not know that we could recover the bicentric ABCD from its
          diagonal triangle EFG.
          I tried to carry out your construction with some minor difficulties for you
          should have labeled the incenter and the circumcenter of the bicentric ABCD
          in your post.
          In this post I label U the incenter, keeping the label O for the center of
          the nine point conic of ABCD just like you.

          .
          >
          > The construction of ABCD now follows by extending the line connecting O
          > with the mid-point E' of FG. One of the points of intersection of this
          > line with the nine-point circle of EFG is the point O of the
          > quadrilateral ABCD we are looking for (the other point of intersection
          > might correspond to an escribable quadrilateral).
          >



          I notice that this line is UE' with my label U for the incenter and your
          > label E' for the midpoint of FG. This line cuts the EFG-nine point circle of
          > course in E' and again in the sought point O, center of the nine point conic
          > of ABCD.
          >




          > The rest of your construction.
          > The rest of your construction is easy.
          >

          When I say easy, of course it is for Cabri users for they can draw any conic
          on 5 points and get their intersection with any other object as lines,
          circles or conics but for GSP lovers it can give them some headaches.
          So, cheerup for your macros and as I have GSP now, thanks to one of my sons
          living in the States, that just the first thing I do, to create a macro
          giving intersection of a conic through 5 given points with a line.
          I suppose there is many books written on GSP and maybe you can recommend me
          one of them?
          Is there a GSP forum just like the Cabri forum?
          Friendly
          Francois


          >


          [Non-text portions of this message have been removed]
        • Eisso J. Atzema
          Dear François, my apologies. I was a bit in a hurry when I wrote up the construction. I looked at the construction again and I noticed something that was new
          Message 4 of 14 , Sep 10, 2007
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            Dear François,

            my apologies. I was a bit in a hurry when I wrote up the construction. I
            looked at the construction again and I noticed something that was new to
            me. It would appear that the circumcircle and the incircle of a
            bicentric quadrilateral determine a non-intersecting coaxal system of
            circles that has the line connecting the midpoints of EF and EG for its
            axis and E as one of the limiting points. This would allow for the
            easy determination of the radius of the incircle in terms of EFG. From
            the fact that R=R[EFG]*sqrt(-cos(E)cos(F)cos(G)), I find that
            r^2=(e^2-f^2-g^2)/2(1+sqrt(tan(F)tan(G))), which expression is confirmed
            numerically by SketchPad. I wonder if this same expression cannot be
            rewritten more symmetrically. Also, do you think a similar construction
            might be given for quadrilaterals which are cyclic and have an excircle?

            Eisso

            --

            ========================================
            Eisso J. Atzema, Ph.D.
            Department of Mathematics & Statistics
            University of Maine
            Orono, ME 04469
            Tel.: (207) 581-3928 (office)
            (207) 866-3871 (home)
            Fax.: (207) 581-3902
            E-mail: atzema@...
            ========================================
          • Francois Rideau
            Dear Eisso Your first formula R=R[EFG]*sqrt(-cos(E)cos(F)cos(G)), is well known. This is the radius of the polar circle of triangle EFG. Inversion wrt this
            Message 5 of 14 , Sep 11, 2007
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              Dear Eisso
              Your first formula
              R=R[EFG]*sqrt(-cos(E)cos(F)cos(G)),
              is well known.
              This is the radius of the polar circle of triangle EFG.
              Inversion wrt this circle swaps EFG-circumcircle and EFG-Euler circle.

              Given 2 circles :<Gamma> with center O and radius R and <gamma> with center
              o and radius r with distance of the centers Oo = d, the existence of a
              bicentric quadrilateral with circumcircle <Gamma> and incircle is ensured by
              the Fuss formula:
              2r^2(R^2 + d^2) = (R^2 - d^2)^2
              Nicolaus Fuss (1755-1826) was a student and a friend of Leonhard Euler who
              gives the formula for the bicentric triangle.
              Fuss also found the formula for the bicentric pentagon, hexagon, heptagon
              and octagon, (Nova Acta Petropol., XIII, 1798).

              Also, do you think a similar construction
              might be given for quadrilaterals which are cyclic and have an excircle?
              In my previous post, I give the formula : d = R for the existence of such a
              quadrilateral but in this case, there is no diagonal triangle for 2 of its 3
              vertices are on the line at infinity!

              Friendly
              Francois


              [Non-text portions of this message have been removed]
            • Francois Rideau
              Dear Eisso I think my answer abour yout last question on the excircle is not very satisfying. Just a word about this bicentric quadrilateral configuration
              Message 6 of 14 , Sep 11, 2007
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                Dear Eisso
                I think my answer abour yout last question on the "excircle" is not very
                satisfying.
                Just a word about this bicentric quadrilateral configuration and I think to
                GSP users who can have some difficulties to draw it.

                The best is to start with the incircle <gamma> with center o and radius r.
                We choose any diameter as axis of symmetry for the coming drawing and on
                this line any point E which have vocation to become the vertice of the
                diagonal triangle located on this axis. Better is to start with E inside
                <gamma> for the beginning.

                Now we choose any point a on circle <gamma>. Line aE cuts again <gamma> in
                point c. Line perpendicular at E to this line aEc cuts circle <gamma> in
                points b and d. Points a, b, c, d are contact points of the bicentric ABCD
                with <gamma>.
                Then it is easy to draw ABCD with A = Td /\ Ta, B = Ta /\ Tb, C = Tb /\ Tc ,
                D = Tc /\ Td where Ta, Tb, Tc, Td are tangent to <gamma> at points a, b, c,
                d. E is a fixed diagonal point for ABCD, the others being F = Ta /\ Tc and G
                = Tb /\Td
                We check that A, B, C, D are on a same circle <Gamma> independant on the
                choice of point a on circle <gamma>.
                Then the drawing is done and from now on, we can look at its properties and
                for example verify the Fuss formula for any position of E on the axis and so
                on....
                If E is inside <gamma>, then <gamma> is an incircle but if E is outside
                <gamma>, <gamma> is an excircle always with the same Fuss formula.
                In this case, E must not be too far from the incenter o in order to line
                perpendicular to line aEc at E can cut circle <gamma> in points b and d. I
                think E must be inside the orthoptic circle of <gamma>.

                In your construction of the bicentric ABCD from the diagonal triangle EFG, I
                think that in general there are 2 solutions, one with an incircle and the
                other with an excircle.
                When you take intersection of circle with diameter FG with the axis OE,
                (sorry I change your labels!), you have 2 intersections o and o' where o and
                O are in the same FG-half plane and o' and E in different FG- half planes.
                With the choice o, we get an incircle and for the choice o' the excircle.
                Now as I have said it, the choice o' don't give always a real solution, so
                there is some discussion on the shape of triangle EFG to see when the o'
                choice gives birth to a real solution.
                Friendly
                François



                On 9/10/07, Eisso J. Atzema <atzema@...> wrote:
                >
                > Dear François,
                >
                > my apologies. I was a bit in a hurry when I wrote up the construction. I
                > looked at the construction again and I noticed something that was new to
                > me. It would appear that the circumcircle and the incircle of a
                > bicentric quadrilateral determine a non-intersecting coaxal system of
                > circles that has the line connecting the midpoints of EF and EG for its
                > axis and E as one of the limiting points. This would allow for the
                > easy determination of the radius of the incircle in terms of EFG. From
                > the fact that R=R[EFG]*sqrt(-cos(E)cos(F)cos(G)), I find that
                > r^2=(e^2-f^2-g^2)/2(1+sqrt(tan(F)tan(G))), which expression is confirmed
                > numerically by SketchPad. I wonder if this same expression cannot be
                > rewritten more symmetrically. Also, do you think a similar construction
                > might be given for quadrilaterals which are cyclic and have an excircle?
                >
                > Eisso
                >
                > --
                >
                > ========================================
                > Eisso J. Atzema, Ph.D.
                > Department of Mathematics & Statistics
                > University of Maine
                > Orono, ME 04469
                > Tel.: (207) 581-3928 (office)
                > (207) 866-3871 (home)
                > Fax.: (207) 581-3902
                > E-mail: atzema@... <atzema%40math.umaine.edu>
                > ========================================
                >
                >
                >


                [Non-text portions of this message have been removed]
              • Eisso J. Atzema
                Dear Hauke, you are completely right that for every (obtuse) triangle there is a bicentric quadrilateral which has this triangle for its diagonal points
                Message 7 of 14 , Sep 11, 2007
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                  Dear Hauke,

                  you are completely right that for every (obtuse) triangle there is a
                  bicentric quadrilateral which has this triangle for its diagonal points
                  triangle. Here is an edited version of my earlier geometric construction.

                  Let EFG be the diagonal triangle of a bicentric quadrilateral ABCD,
                  where E is the point of intersection of the diagonals AC and BD. Since
                  ABCD is to be cyclic, we need to assume that the angle FEG is obtuse.
                  For the same reason, ABCD is assumed to be convex. Now:

                  1) Note that for every complete quadrilateral ABCD its diagonal triangle
                  EFG is self-polar with respect to every conic circumscribing ABCD. If ABCD
                  is cyclic, that means that EFG has to be self-polar w.r.t. a circle. But
                  a necessary condition for a triangle to be self-polar w.r.t. a circle is
                  that the center of the circle is the orthocenter of the triangle. This
                  determines the position of the circumcenter Z of ABCD. The radius of the
                  circle is easily determined to be equal to
                  2R[EFG]*sqrt(-cos(E)cos(F)cos(G)).

                  2) If ABCD is to be cyclic, then its nine-point conic is an orthogonal
                  hyperbola which has its center at the center point O of the quadrilateral.

                  a. It is a known theorem that for ABCD cyclic, O lies on the nine-point
                  circle of EFG. This means that the locus of all possible center points
                  is well-determined.

                  b. For any orthogonal hyperbola, conjugation is reflection in its
                  asymptotes. Since for every quadrilateral the diameters of its
                  nine-point parallel to a pair of opposite sides are conjugate, it
                  follows that the directions of the three pairs of opposite sides of a
                  cyclic ABCD are obtained by reflection in the asymptotes of the
                  nine-point conic. Particularly, this means that the angle bisectors at
                  E, F and G are all parallel to the asymptotes of the nine-point conic.

                  3) If ABCD is to be inscribable as well, the interior angle bisectors at
                  F and G need to meet at the incenter of the quadrilateral. Since these
                  two angle bisectors are perpendicular by 2b, it follows that the
                  incenter lies on the circle that has the long side FG of EFG for its
                  diameter. Moreover, it is known for any bicentric quadrilateral, its
                  incenter lies on the line connecting its diagonal point E to its
                  circumcenter. This completely determines the position of the incenter I of
                  any bicentric quadrilateral ABCD which has EFG for its diagonal triangle.

                  4) The construction of ABCD follows by noting that I also lies on the
                  line connecting the midpoints of AC and BD since this line is the locus
                  of all points P such that |ABP|+|CDP| =|ADP|+|BCP|. For the same reason,
                  the midpoint of FG is on the same line (aka Gauss-Bodenmiller). We can
                  now find O by extending the line connecting I with the mid-point E' of
                  FG. One of the points of intersection of this
                  line with the nine-point circle of EFG is the point O of the
                  quadrilateral ABCD we are looking for. Since O is the center point of
                  ABCD, it is also the center of the ninepoint conic of ABCD. Since E,F,G
                  (as well as Z) are on the
                  same nine-point conic, this conic is completely determined. The six
                  mid-points of the sides of ABCD are now found by noting that any
                  bimedian of ABCD bisects the corresponding side of EFG (extended
                  Gauss-Bodenmiller). Since these mid-points are also on the nine-point
                  conic, they are the points of intersections of the lines
                  connecting O with the mid-points of EFG. Once we have the mid-points of
                  ABCD, we find the quadrilateral itself by connecting E, F, and G with
                  the corresponding mid-points. This works out nicely without any
                  ambiguity and we have our bicentric quadrilateral ABCD.

                  As for the second point of intersection of ZI with the circle that has
                  FG for a diameter, it would appear that generally it corresponds to an
                  imaginary quadrilateral. The corresponding nine-point conic and the
                  lines connecting the midpoints of EGF with the corresponding point O do
                  not all meet. Most likely this expresses that there cannot be a
                  quadrilateral that is both cyclic and has an escribed circle.

                  The whole construction can be done directly in Cabri. If you are using
                  SketchPad, you will want to have some macros to draw a conic section and
                  to find the points of intersection of a conic section with a straight line.

                  Eisso



                  --

                  ========================================
                  Eisso J. Atzema, Ph.D.
                  Department of Mathematics & Statistics
                  University of Maine
                  Orono, ME 04469
                  Tel.: (207) 581-3928 (office)
                  (207) 866-3871 (home)
                  Fax.: (207) 581-3902
                  E-mail: atzema@... <mailto:atzema%40math.umaine.edu>
                  ========================================
                • Hauke Reddmann
                  ... using ... section and ... straight line. ... Could you do the actual construction for the Kimberling triangle? This way one can easily check U (=#4?) and
                  Message 8 of 14 , Sep 16, 2007
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                    >
                    > The whole construction can be done directly in Cabri. If you are
                    using
                    > SketchPad, you will want to have some macros to draw a conic
                    section and
                    > to find the points of intersection of a conic section with a
                    straight line.
                    >
                    Could you do the actual construction for the Kimberling
                    triangle? This way one can easily check U (=#4?) and nail
                    down I.

                    Here are the edges:

                    a:=6;b:=9;c:=13;
                    s:=(a+b+c)/2;f:=sqrt(s*(s-a)*(s-b)*(s-c));
                    ax:=(a**2+b**2-c**2)/2/a;ay:=2*f/a;bx:=a;by:=0;cx:=0;cy:=0;

                    It's only these values to check, so no macros needed, I guess.

                    Hauke
                  • Francois Rideau
                    Dear Eisso In fact I knew the coaxal properties of both incircle and circumcircle of the bicentric ABCD for line FG is the polar line of E wrt both circles
                    Message 9 of 14 , Sep 17, 2007
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                      Dear Eisso

                      In fact I knew the coaxal properties of both incircle and circumcircle of
                      the bicentric ABCD

                      for line FG is the polar line of E wrt both circles leading to a better and
                      shorter construction of the bicentric ABCD given its diagonal triangle EFG
                      but I was so astonished by all your super subtle idea that I forgot it!



                      I come back now to the discussion of your construction.

                      The circumcenter is known as the orthocenter of triangle EFG and to have
                      reality of the circumcircle, angle <E must be obtuse, OK.

                      Now the circle of diameter FG cuts line on E perpendicular to FG in 2 points
                      I and I' candidates to be the incenter.

                      Point I located in the FG- half-plane containing E gives always birth to a
                      real solution but the other point I' located in the other FG-half-plane can
                      give no real solution.

                      I have found that in order to I' -choice leads to a real solution, both
                      following inequalities must be satisfied by angles <F and <G of triangle
                      EFG:

                      tan(<F)^3 < tan(<G)

                      and

                      tan(<G)^3 < tan(<F)

                      That sounds weird to me !

                      So triangles EFG such that :

                      tan(<F)^3 = tan(<G)

                      or

                      tan(<G)^3 = tan(<F)

                      are very special limiting triangles for your problem.

                      It is easy to construct them with Cabri or GSP, for example by using the
                      function x --> Arctan(tan(x)^3) or by any more direct (ruler-compass) means
                      suggested by this problem that I let you find.



                      Friendly

                      Francois


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