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Re: [EMHL] Squares inscribed in triangle

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  • Antreas P. Hatzipolakis
    ... What ??? I didn t send any non-text or has my mailer itself sent it ? APH
    Message 1 of 16 , Dec 1, 2001
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      On Saturday, December 1, 2001, at 07:52 AM, Hyacinthos wrote:

      >
      > [Non-text portions of this message have been removed]
      >
      >

      What ??? I didn't send any non-text or has my mailer itself sent it ?


      APH
    • jean-pierre.ehrmann@wanadoo.fr
      Dear Atul and other Hyacinthists ... sides of ... Let P =(x,y,z) in barycentrics. As the length of the sides of the square inscribed in ABC is 2ad/(aa+2d)where
      Message 2 of 16 , Dec 2, 2001
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        Dear Atul and other Hyacinthists

        > And also is there a point P for which the lengths of the
        sides of
        > inscribed squares in each of the triangles PBC,PCA and PAB equal?


        Let P =(x,y,z) in barycentrics.
        As the length of the sides of the square inscribed in ABC is
        2ad/(aa+2d)where d=area(ABC), the length of the sides of a square
        inscribed in PBC is 2dax/(aa(x+y+z)+2dx).
        The three length are equal iff
        u = v = w = 0
        where u = bc(cy-bz)(x+y+z)+2d(b-c)yz; v,w circular.
        Now, as 1/(2d)(aau+bbv+ccw)=aa(b-c)yz+bb(c-a)zx+cc(a-b)xy, P lies
        necessarily on the circumhyperbola through K and I.
        Putting x=aa/(a-t), y=..., we get
        (E) : aa/(t-a)+bb/(t-b)+cc/(t-c)+2d/t = 0.
        If we suppose, for example, a<b<c, (E) has three real roots,
        one in ]0,a[, one in ]a,b[, one in ]b,c[.
        Only the first one gives a point inside ABC and t is the common value
        of the lengths of the sides of the three squares.

        Conclusion :
        There exists an unique point P inside ABC for which the lengths of
        the sides of the three squares are equal; the common length is the
        smallest root t of (E) and P is barycentric(aa/(a-t),...).
        P is not constructible and lies on the circumhyperbola through I and
        K.

        Note that we easily find a conic-construction of P, using the three
        common points of the mesh generated by u=0, v=0, w=0.

        Another interesting problem could be :
        If Pa,Pb,Pc are the centers of the three squares inscribed in PBC,
        PCA, PAB, find the locus of P for which PaPbPc and ABC are
        perspective.
        This locus is, if I didn't mistake, the self-isogonal circumcubic
        with pivot U, where
        Vect(KU) = -4d/(aa+bb+cc) Vect(OH).
        I think that this point U can be interesting for Clark because he
        lies on the parallel through K to the Euler line and it is the first
        time that I meet a triangle center with an interesting geometric
        property on this line (except, of course, K and the infinite point).

        Of course, putting -d instead of d, we get similar results for the
        the three other squares inscribed in the side lines of PBC,...

        I hope that there are not too many mistakes.
        Friendly. Jean-Pierre
      • xpolakis@otenet.gr
        ... rectangles inscribed ... whose ... ^^^^ T ... of ... ^^^^ t ... Antreas
        Message 3 of 16 , Dec 2, 2001
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          --- In Hyacinthos@y..., "Antreas P. Hatzipolakis" <xpolakis@o...>
          wrote:
          > Let ABC be a triangle and P = (x:y:z) a point in normals.
          > Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR
          rectangles inscribed
          > in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp.,
          whose
          > base / height = t.
          ^^^^
          T

          >
          > A
          > /\
          > / \ AbAc / AbA'b =T
          > / \
          > / \
          > / \
          > / P \
          > / \
          > / A'b M'a A'c \
          > / Pa \
          > B----Ab--Ma--Ac----C
          >
          >
          > Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c
          of
          > AbAcA'cA'b,
          > and similarly MbM'b, McM'c.
          >
          > Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp.
          such that:
          >
          > M'aPa M'bPb M'cPc
          > ----- = ----- = ----- = T
          > M'aMa M'bMb M'cMc

          ^^^^
          t


          >
          >
          > ABC, PaPbPc are perspective <==>
          >
          > (1+t)cosB + TsinB
          > x(-------------------) + z
          > 2
          > --------------------------- * CYCLIC = 1
          > (1+t)cosC + TsinC
          > x(-------------------) + y
          > 2
          >
          > ==> The locus of P is the isogonal pivotal cubic with pivot in
          normals:
          >
          > (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
          > ((------------------) - [(------------------) * (------------------)] ::)
          > 2 2 2
          >
          > The case of the squares and their centers is for T = 1, t = 1/2.

          Antreas
        • Antreas P. Hatzipolakis
          Dear Jean-Pierre, [JPE] ... -- I think that we have seen this problem before, and also its parametrizations. Let ABC be a triangle and P = (x:y:z) a point in
          Message 4 of 16 , Dec 2, 2001
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            Dear Jean-Pierre,

            [JPE]
            >Another interesting problem could be :
            >If Pa,Pb,Pc are the centers of the three squares inscribed in PBC,
            >PCA, PAB, find the locus of P for which PaPbPc and ABC are
            >perspective.
            >This locus is, if I didn't mistake, the self-isogonal circumcubic
            >with pivot U, where
            >Vect(KU) = -4d/(aa+bb+cc) Vect(OH).
            >I think that this point U can be interesting for Clark because he
            >lies on the parallel through K to the Euler line and it is the first
            >time that I meet a triangle center with an interesting geometric
            >property on this line (except, of course, K and the infinite point).
            >
            >Of course, putting -d instead of d, we get similar results for the
            >the three other squares inscribed in the side lines of PBC,...

            --
            I think that we have seen this problem before, and also its
            parametrizations.

            Let ABC be a triangle and P = (x:y:z) a point in normals.
            Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles inscribed
            in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
            base / height = t.

            A
            /\
            / \ AbAc / AbA'b = t
            / \
            / \
            / \
            / P \
            / \
            / A'b M'a A'c \
            / Pa \
            B----Ab--Ma--Ac----C


            Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
            AbAcA'cA'b,
            and similarly MbM'b, McM'c.

            Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such that:

            M'aPa M'bPb M'cPc
            ----- = ----- = ----- = T
            M'aMa M'bMb M'cMc


            ABC, PaPbPc are perspective <==>

            (1+t)cosB + TsinB
            x(-------------------) + z
            2
            --------------------------- * CYCLIC = 1
            (1+t)cosC + TsinC
            x(-------------------) + y
            2

            ==> The locus of P is the isogonal pivotal cubic with pivot in normals:

            (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
            ((------------------) - [(------------------) * (------------------)] ::)
            2 2 2

            The case of the squares and their centers is for T = 1, t = 1/2.

            Quoting you but applying it for me:

            >I hope that there are not too many mistakes.


            Greetings from Athens

            Antreas
          • Antreas P. Hatzipolakis
            Let ABC be a triangle, and AbAcA cA b, BcBaB cB a, CaCbC bC a the three similar rectangles inscribed in ABC, based on BC, CA, AB, resp., whose Height / Base =
            Message 5 of 16 , Dec 2, 2001
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              Let ABC be a triangle, and AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a
              the three similar rectangles inscribed in ABC, based on BC, CA,
              AB, resp., whose Height / Base = t.
              Let AA', BB', CC' be the three altitudes of ABC.



              A
              /\
              / \ A'bAb / AbAc = t
              / \
              / \
              / \
              A'b A'c
              / \
              / \
              / \
              B---Ab----A'--Ac---C

              If (1/AbAc) + (1/BcBa) + (1/CaCb) = (1/AA') + (1/BB') + (1/CC')
              then find t.

              Antreas

              PS: The t that I found equals to:

              sinBsinC + sinCsinA + sinAsinB
              1 - ( ------------------------------ )
              sinA + sinB + sinC

              APH
            • jean-pierre.ehrmann@wanadoo.fr
              Dear Antreas ... inscribed ... )] ::) ... I m sorry but I didn t remember this discussion. Note that the locus of the pivot is the hyperbola through G, the
              Message 6 of 16 , Dec 3, 2001
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                Dear Antreas

                > Let ABC be a triangle and P = (x:y:z) a point in normals.
                > Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles
                inscribed
                > in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
                > base / height = t.
                >
                > A
                > /\
                > / \ AbAc / AbA'b = t
                > / \
                > / \
                > / \
                > / P \
                > / \
                > / A'b M'a A'c \
                > / Pa \
                > B----Ab--Ma--Ac----C
                >
                >
                > Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
                > AbAcA'cA'b,
                > and similarly MbM'b, McM'c.
                >
                > Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such
                that:
                >
                > M'aPa M'bPb M'cPc
                > ----- = ----- = ----- = T
                > M'aMa M'bMb M'cMc
                >
                >
                > ABC, PaPbPc are perspective <==>
                >
                > (1+t)cosB + TsinB
                > x(-------------------) + z
                > 2
                > --------------------------- * CYCLIC = 1
                > (1+t)cosC + TsinC
                > x(-------------------) + y
                > 2
                >
                > ==> The locus of P is the isogonal pivotal cubic with pivot in
                normals:
                >
                > (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
                > ((------------------) - [(------------------) * (------------------
                )] ::)
                > 2 2 2
                >
                > The case of the squares and their centers is for T = 1, t = 1/2.

                I'm sorry but I didn't remember this discussion.
                Note that the locus of the pivot is the hyperbola through G, the
                Steiner point, the Fermat points and the reflection of G wrt O
                (surely, this has been discussed too).

                If we take P inside ABC and lying on the circumhyperbola through I
                and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
                How can we construct these three rectangles?

                I have another question related to the following problem :
                Find P inside ABC such as the incircles of PBC, PCA, PAB are
                congruent.
                I presume that this problem has been discussed and solved (it doesn't
                seem to be not an easy one)
                Have you any reference?

                Friendly. Jean-Pierre
              • Atul Dixit
                Dear sir ... It s a special case of Elkies point and has been named as Congruent incircles point References:
                Message 7 of 16 , Dec 3, 2001
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                  Dear sir

                  >I have another question related to the following problem :
                  >Find P inside ABC such as the incircles of PBC, PCA, PAB are
                  >congruent.
                  >I presume that this problem has been discussed and solved (it doesn't
                  >seem to be not an easy one)
                  >Have you any reference?
                  >Friendly. Jean-Pierre

                  It's a special case of Elkies point and has been named as "Congruent
                  incircles point"

                  References:
                  1)http://mathworld.wolfram.com/CongruentIncirclesPoint.html
                  2)Kimberling, C. "Central Points and Central Lines in the Plane of a
                  Triangle." Math. Mag. 67, 163-187, 1994.

                  Yours faithfully
                  Atul.A.Dixit






















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                • jean-pierre.ehrmann@wanadoo.fr
                  Dear Atul ... doesn t ... as Congruent ... a ... There is nothing in the Weinsstein CRC except the definition and I don t have Clark s article. What I wanted
                  Message 8 of 16 , Dec 3, 2001
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                    Dear Atul
                    > >I have another question related to the following problem :
                    > >Find P inside ABC such as the incircles of PBC, PCA, PAB are
                    > >congruent.
                    > >I presume that this problem has been discussed and solved (it
                    doesn't
                    > >seem to be not an easy one)
                    > >Have you any reference?
                    > >Friendly. Jean-Pierre
                    >
                    > It's a special case of Elkies point and has been named
                    as "Congruent
                    > incircles point"
                    >
                    > References:
                    > 1)http://mathworld.wolfram.com/CongruentIncirclesPoint.html
                    > 2)Kimberling, C. "Central Points and Central Lines in the Plane of
                    a
                    > Triangle." Math. Mag. 67, 163-187, 1994.


                    There is nothing in the Weinsstein CRC except the definition and I
                    don't have Clark's article.
                    What I wanted to know is :
                    Can we find a trigonometric expression of the coordinates of the
                    point?
                    What is the algebraic equation (wrt a,b,c) of lowest degree with root
                    the common radius ? Is there a known curve going through this point?
                    Friendly. Jean-Pierre
                  • jean-pierre.ehrmann@wanadoo.fr
                    Dear Antreas: ... PAB. ... [APH] ... I mean congruent = with both equal bases and equal heights Friendly. Jean-Pierre
                    Message 9 of 16 , Dec 3, 2001
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                      Dear Antreas:

                      [JPE]:
                      > > If we take P inside ABC and lying on the circumhyperbola through I
                      > > and K, we can inscribe three congruent rectangles in PBC, PCA,
                      PAB.
                      > > How can we construct these three rectangles?
                      > >
                      >

                      [APH]
                      > Do you mean congruent = with both equal bases and equal heights or
                      > do you mean equal [in area] ?
                      > That is, having their bases inversely proportional to their heights.

                      I mean congruent = with both equal bases and equal heights

                      Friendly. Jean-Pierre
                    • jean-pierre.ehrmann@wanadoo.fr
                      Dear Antreas and other Hyacinthists ... through I ... heights. [JPE] ... May be, this needs a little explanation. I ve discovered (but I guess that I m not the
                      Message 10 of 16 , Dec 3, 2001
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                        Dear Antreas and other Hyacinthists

                        > [JPE]:
                        > > > If we take P inside ABC and lying on the circumhyperbola
                        through I
                        > > > and K, we can inscribe three congruent rectangles in PBC, PCA,
                        > PAB.
                        > > > How can we construct these three rectangles?
                        > > >
                        > >
                        >
                        > [APH]
                        > > Do you mean congruent = with both equal bases and equal heights or
                        > > do you mean equal [in area] ?
                        > > That is, having their bases inversely proportional to their
                        heights.

                        [JPE]
                        > I mean congruent = with both equal bases and equal heights


                        May be, this needs a little explanation.
                        I've discovered (but I guess that I'm not the first one) that when
                        I've looked at congruent inscribed squares (Recent problem from Atul
                        Dixit)

                        Let m = height/basis.
                        The length of the basis of such a rectangle inscribed in ABC (with
                        basis on BC ) is 2ad/(maa+2d) where d = area(ABC).
                        Thus the length of the basis of the rectangle inscribed in PBC (for a
                        given ratio m) is
                        La = 2adx/(maa(x+y+z)+2dx) if p is barycentric (x,y,z).
                        We have La = Lb = Lc iff
                        u = v = w = 0 where
                        u = bcm(x+y+z)(cy-bz)+2(b-c)dyz; v and w : circular.
                        As 1/(2d)(aau+bbv+ccw) = aa(b-c)yz+bb(c-a)zx+cc(a-b)xy,
                        necessarily P lies on the circumhyperbola through I and K.
                        Putting P = (aa/(a-L),bb/(b-L),cc/(c-L)), we have u=v=w=0 iff
                        (E) : m(aa/(a-L)+bb/(b-L)+cc/(c-L))=2d/L
                        This way, we get three congruent rectangles inscribed in PBC, PCA,
                        PAB with common basis L and common height mL, where m is given by (E).

                        Now, the problem is : find an easy construction of the three
                        rectangles for given P.
                        May be, it can be useful to note that
                        L = 1/3(a+b+c)(IP*/GP*) where P*=isogonal(P) - on the line IG -

                        Friendly. Jean-Pierre
                      • jean-pierre.ehrmann@wanadoo.fr
                        Dear Antreas, ... PAB. ... [APH] ... If P = I, the rectangles degenerate in three segments (basis = 0, height = r) Friendly. Jean-Pierre
                        Message 11 of 16 , Dec 3, 2001
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                          Dear Antreas,

                          > [JPE]:
                          > > If we take P inside ABC and lying on the circumhyperbola through I
                          > > and K, we can inscribe three congruent rectangles in PBC, PCA,
                          PAB.
                          > > How can we construct these three rectangles?
                          > >

                          [APH]
                          >
                          > A
                          > /\
                          > / \
                          > / \
                          > / \
                          > / \
                          > / I \
                          > / A'b A'c\
                          > / \
                          > B---Ab-------Ac--C
                          >
                          > Assume that P = I.
                          >
                          > Let AbAcA'cA'b, BcBaB'aB'c, CaCbC'bC'a be the three
                          > congruent rectangles inscribed in IBC, ICA, IAB, resp.
                          > (based on BC, CA, AB, resp.)
                          >
                          > Denote AbAc = BcBa = CaCb = f
                          >
                          > AbA'b = AcA'c = BcB'c = BaB'a = CaC'a = CbC'b = g
                          >
                          > We have:
                          >
                          > BC = BAb + AbAc + AcC = gcot(B/2) + f + gcot(C/2) or
                          >
                          > a = f + g(cot(B/2) + cot(C/2))
                          >
                          > Similarly:
                          >
                          > b = f + g(cot(C/2) + cot(A/2))
                          >
                          > c = f + g(cot(A/2) + cot(B/2))
                          >
                          > Hmmm.... What do we get from this?


                          If P = I, the rectangles degenerate in three segments (basis = 0,
                          height = r)
                          Friendly. Jean-Pierre
                        • Antreas P. Hatzipolakis
                          Dear Jean-Pierre, [APH]: [parameters corrected] ... Probably yes, probably no! I don t remember. ... Do you mean congruent = with both equal bases and equal
                          Message 12 of 16 , Dec 3, 2001
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                            Dear Jean-Pierre,

                            [APH]:
                            [parameters corrected]
                            >> Let ABC be a triangle and P = (x:y:z) a point in normals.
                            >> Let AbAcA'cA'b, BcBaB'cB'a, CaCbC'bC'a three SIMILAR rectangles
                            >> inscribed
                            >> in the triangles PBC, PCA, PAB, based on BC, CA, AB, resp., whose
                            >> base / height = T.
                            >>
                            >> A
                            >> /\
                            >> / \ AbAc / AbA'b = T
                            >> / \
                            >> / \
                            >> / \
                            >> / P \
                            >> / \
                            >> / A'b M'a A'c \
                            >> / Pa \
                            >> B----Ab--Ma--Ac----C
                            >>
                            >>
                            >> Let now MaM'a be the perp. bisector of the bases AbAc, A'bA'c of
                            >> AbAcA'cA'b,
                            >> and similarly MbM'b, McM'c.
                            >>
                            >> Let Pa, Pb, Pc be three points on MaM'a, MbM'b, McM'c, resp. such that:
                            >>
                            >> M'aPa M'bPb M'cPc
                            >> ----- = ----- = ----- = t
                            >> M'aMa M'bMb M'cMc
                            >>
                            >>
                            >> ABC, PaPbPc are perspective <==>
                            >>
                            >> (1+t)cosB + TsinB
                            >> x(-------------------) + z
                            >> 2
                            >> --------------------------- * CYCLIC = 1
                            >> (1+t)cosC + TsinC
                            >> x(-------------------) + y
                            >> 2
                            >>
                            >> ==> The locus of P is the isogonal pivotal cubic with pivot in normals:
                            >>
                            >> (1+t)cosA + TsinA (1+t)cosB + TsinB (1+t)cosC + TsinC
                            >> ((------------------) - [(------------------) * (------------------)]
                            >> ::)
                            >> 2 2 2
                            >>
                            >> The case of the squares and their centers is for T = 1, t = 1/2.
                            >

                            [JPE]:
                            > I'm sorry but I didn't remember this discussion.
                            > Note that the locus of the pivot is the hyperbola through G, the
                            > Steiner point, the Fermat points and the reflection of G wrt O
                            > (surely, this has been discussed too).

                            Probably yes, probably no! I don't remember.

                            [JPE]:
                            >
                            > If we take P inside ABC and lying on the circumhyperbola through I
                            > and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
                            > How can we construct these three rectangles?
                            >

                            Do you mean congruent = with both equal bases and equal heights or
                            do you mean equal [in area] ?
                            That is, having their bases inversely proportional to their heights.

                            [JPE]:
                            > I have another question related to the following problem :
                            > Find P inside ABC such as the incircles of PBC, PCA, PAB are
                            > congruent.
                            > I presume that this problem has been discussed and solved (it doesn't
                            > seem to be not an easy one)
                            > Have you any reference?
                            >

                            Let P = (x:y:z) in actual normals.

                            Let R_a, R_b, R_c be the inradii of the triangles PBC, PCA, PAB.

                            R_a = R_b = R_c <==>

                            area(PBC)/semiper.(PBC) = area(PCA)/semiper.(PCA) =
                            area(PAB)/semiper.(PAB)

                            <==> ax/(a + PB + PC) = by/(b + PC + PA) = cz/(c + PA + PB) <==>

                            xsinA/(a + PB + PC) = ysinB/(b + PC + PA) = zsinC/(c + PA + PB) [1]

                            Now, we have a = (xsinA + ysinB + zsinC)/sinBsinC, b = ..., c = ...

                            sqrt(y^2 + z^2 + 2yzcosA)
                            PA = -------------------------, PB = ..., PC =....
                            sinA

                            The system [1] now can be homogenized but, as Paul would say, neither I
                            nor my computer could solve it!


                            Antreas


                            [Non-text portions of this message have been removed]
                          • Antreas P. Hatzipolakis
                            ... For a plane triangle A1A2A3, call two circles within the triangle companion incircles if they are the incircles of two triangles formed by dividing
                            Message 13 of 16 , Dec 3, 2001
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                              [Atul Dixit]:
                              > It's a special case of Elkies point ...

                              For a plane triangle A1A2A3, call two circles within the triangle
                              <i>companion incircles</i> if they are the incircles of two triangles
                              formed by dividing A1A2A3 into two smaller triangles by passing a line
                              through one of the vertices and some point on the opposite side.
                              (a) Show that any chain of circles C1,C2,... such that Ci, Ci+1 are
                              companion incircles for every i consists of at most six distinct circles.
                              (b) Give a ruler and compass construction for the unique chain which
                              has only three distinct circles.
                              (c) For such a chain of three circles show that the three subdividing
                              lines are concurrent.
                              (The American Mathematical Monthly 94(1987) 877 by Noam Elkies)

                              Antreas
                            • Antreas P. Hatzipolakis
                              Dear Jean-Pierre, ... A / / / / / / I / A b A c / B---Ab-------Ac--C Assume that P = I. Let AbAcA cA b,
                              Message 14 of 16 , Dec 3, 2001
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                                Dear Jean-Pierre,

                                [JPE]:
                                > If we take P inside ABC and lying on the circumhyperbola through I
                                > and K, we can inscribe three congruent rectangles in PBC, PCA, PAB.
                                > How can we construct these three rectangles?
                                >

                                A
                                /\
                                / \
                                / \
                                / \
                                / \
                                / I \
                                / A'b A'c\
                                / \
                                B---Ab-------Ac--C

                                Assume that P = I.

                                Let AbAcA'cA'b, BcBaB'aB'c, CaCbC'bC'a be the three
                                congruent rectangles inscribed in IBC, ICA, IAB, resp.
                                (based on BC, CA, AB, resp.)

                                Denote AbAc = BcBa = CaCb = f

                                AbA'b = AcA'c = BcB'c = BaB'a = CaC'a = CbC'b = g

                                We have:

                                BC = BAb + AbAc + AcC = gcot(B/2) + f + gcot(C/2) or

                                a = f + g(cot(B/2) + cot(C/2))

                                Similarly:

                                b = f + g(cot(C/2) + cot(A/2))

                                c = f + g(cot(A/2) + cot(B/2))

                                Hmmm.... What do we get from this?


                                Antreas


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