Loading ...
Sorry, an error occurred while loading the content.
 

Concurrencies on a trianle.

Expand Messages
  • Kostas Vittas
    Dear all my friends. I would like to present an interesting configuration as follows : ( I apologize for my English ) A triangle ABC is given and let AD, BE,
    Message 1 of 4 , Jun 1, 2006
      Dear all my friends.

      I would like to present an interesting configuration as follows :
      ( I apologize for my English )

      A triangle ABC is given and let AD, BE, CF be, it's altitudes.
      We draw three circles (K1), (K2), (K3), with side segments BC, AC, AB
      respectively, as diameters.

      We denote as D', D'', the intersection points of line AD from
      circle (K1). Similarly we denote the pair of points E', E'' and F',
      F'', on lines BE, CF, respectively. ( The points D', E', F', inwardly
      to the triangle ABC ).

      We denote A', as the intersection point of the lines E'F'',
      E''F'. Similarly B' as the one of the lines D'F'', D''F' and C' as
      the one of the lines D'E'', D''E'.

      Results :

      1. The points A', B', C', lie on BC, AC, AB, respectively.
      2. The lines AA', BB', CC', are concurrent at a point, so be it M.
      3. The lines A'D',B'E',C'F', are concurrent at a point, so be it L.
      4. The lines A'D'',B'E'',C'F'',are concurrent at a point,so be it N.
      5. The points L, M, N, lie on a line pass through orthocenter H, of
      the triangle ABC.
      6. The points L, N, are the circumcenters of the triangles D'E'F',
      D''E''F'', respectively.
      7. If D1, E1, F1, are the intersection points of circles (K1), (K2),
      (K3) respectively, from the circumcircle of the triangle
      D''E''F'', then the lines AD1, BE1, CF1, are concurrent.
      8. If D2, E2, C2, are the intersection points of circles (K1), (K2),
      (K3) respectively, from the circumcircle of the triangle D'E'F',
      then the lines D1D2, E1E2, F1F2, are concurrent at orthocenter H
      of ABC.

      I will try to post here next time, some proofs I have in mind for
      (1), (2),...(6). I still not have proofs for (7), (8) and I am
      waiting, with pleasure, for any idea. ( My drawing is erected on a
      triangle ABC such that BC = 9.5, AC = 8.5, AB = 6.5 ).

      Best regards.
      Kostas Vittas.
    • Quang Tuan Bui
      Dear Kostas, You can see some results (only true if ABC is acute): E E F F are concyclic on circle (T1) centered at A and orthogonal with (K1), cutting (K1)
      Message 2 of 4 , Jun 1, 2006
        Dear Kostas,
        You can see some results (only true if ABC is acute):
        E'E''F'F'' are concyclic on circle (T1) centered at A and orthogonal with (K1), cutting (K1) at X1, X2. Similarly we have (T2) and Y1, Y2; (T3) and Z1, Z2. Six points X1, X2, Y1, Y2, Z1, Z2 are concyclic on one circle centered at centroid G of triangle ABC.
        Best regards,
        Bui Quang Tuan

        Kostas Vittas <vittasko@...> wrote:
        ...
        A triangle ABC is given and let AD, BE, CF be, it's altitudes.
        We draw three circles (K1), (K2), (K3), with side segments BC, AC, AB
        respectively, as diameters.

        We denote as D', D'', the intersection points of line AD from
        circle (K1). Similarly we denote the pair of points E', E'' and F',
        F'', on lines BE, CF, respectively. ( The points D', E', F', inwardly
        to the triangle ABC ).

        ...

        ---------------------------------
        Yahoo! Messenger with Voice. Make PC-to-Phone Calls to the US (and 30+ countries) for 2¢/min or less.

        [Non-text portions of this message have been removed]
      • Kostas Vittas
        ... N. ... Dear all my friends. I would like to present the proof of proposition (7), as follows: ( my drawing is erected on a triangle with sides BC = 7.5,
        Message 3 of 4 , Jun 13, 2006
          --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
          wrote:

          > Dear all my friends.
          >
          > I would like to present an interesting configuration as
          >follows : ( I apologize for my English )
          >
          > A triangle ABC is given and let AD, BE, CF be, it's altitudes.
          > We draw three circles (K1), (K2), (K3), with side segments BC, AC,
          >AB respectively, as diameters.
          >
          > We denote as D', D'', the intersection points of line AD from
          > circle (K1). Similarly we denote the pair of points E', E'' and F',
          > F'', on lines BE, CF, respectively. ( The points D', E', F',
          >inwardly to the triangle ABC ).
          >
          > We denote A', as the intersection point of the lines E'F'',
          > E''F'. Similarly B' as the one of the lines D'F'', D''F' and C' as
          > the one of the lines D'E'', D''E'.
          >
          > Results :
          >
          > 1. The points A', B', C', lie on BC, AC, AB, respectively.
          > 2. The lines AA', BB', CC', are concurrent at a point, so be it M.
          > 3. The lines A'D',B'E',C'F', are concurrent at a point, so be it L.
          > 4. The lines A'D'',B'E'',C'F'',are concurrent at a point,so be it
          N.
          > 5. The points L, M, N, lie on a line pass through orthocenter H,
          > of the triangle ABC.
          > 6. The points L, N, are the circumcenters of the triangles D'E'F',
          > D''E''F'', respectively.
          > 7. If D1, E1, F1, are the intersection points of circles (K1),
          > (K2),(K3) respectively, from the circumcircle of the triangle
          > D''E''F'', then the lines AD1, BE1, CF1, are concurrent.
          > 8. If D2,E2,F2, are the intersection points of circles (K1),(K2),
          > (K3)respectively, from the circumcircle of the triangle D'E'F',
          > then the lines D1D2, E1E2, F1F2, are concurrent at orthocenter
          > H of ABC.
          >
          > ...

          Dear all my friends.

          I would like to present the proof of proposition (7), as
          follows: ( my drawing is erected on a triangle with sides
          BC = 7.5, AC = 6.8, AB = 5.2 )

          7a. Let A' be the intersection point of E1E'',F1F'' and B', the
          intersection point of D1D'', F1F''and C',the intersection point
          of D1D'', E1E''.

          7b. The point A' is the radical center of the circles (K2),(K3),(M)
          ( the circumcircle of the triangle D''E''F'' ). That is the
          segment line AD, passes through the point A'. Similarly the
          segment lines BE, CF, pass through the points B',C'respectively.

          7c. Let R be the intersection point of BC, D1D''. We draw a line
          perpendicular to the segment line BC at point R, which cuts the
          segment lines BD1, CD1, at points Q, S respectively.

          7d. The point C, is the orthocenter of the triangle BQS ( because of
          the lines BR,SD1,are perpendicular to the lines QS,BQ
          respectively ).Hence the segment lines QC, BS,are perpendicular
          each other and their intersection point, so be it D'1, lies on
          the circle (K1).

          7e. We will prove that the point D'1 coincides to the point D'.
          Really from the cyclic quadrilateral CD1QR we have that
          <RCQ = <RD1Q (1). But <RCQ = <BCD'1 (2) and <RD1Q = <BD1D''(3).
          From (1),(2),(3) => <BCD'1 = <BD1D'' => BD''= BD'1 (4)
          Hence the point D'1 coincides to the point D'. That is, we have
          that the segment line CD' passes through the point Q.

          7f. As a helping proposition, we use the result that the points B',
          A, S, are collinear ( we shall prove the proposition for this
          result, below ). Hence, we consider the segments AD', BB' CC',
          concurrent at orthocenter H of ABC and then, by the Lemma 2, we
          have that the points Q' ( the intersection point of AC', CD' ),
          R, S, as the intersection points of the opposite sides, the
          other than H, of the quadrilaterals, AD'CC', BCC'B', AB'BD'
          respectively, are collinear. Hence the point Q' coincides to
          the point Q and then we have that the segment line AC', passes
          through the point Q.

          7g. Now, because of the points Q, R, S, are collinear, we conclude
          that the triangles AB'C', BCD1, are perspective at one point, so
          be it A1. That is the point A1, as the intersection point of
          BC',B'C,lies on AD1. Similarly The point B1, as the intersection
          point of AC',A'C and the point C1, as the intersection point of
          AB', A'B, lie on BE1, CF1 respectively.

          7h. We consider the segments AA',BB',CC', concurrent at orthocenter
          H of ABC. So, by the Lemma 1, we have that the lines AA1, BB1,
          CC1, are concurrent. Hence the lines AD1, BE1, CF1, are
          concurrent at one point, so be it P and the proof of proposition
          (7), is completed.

          7.1 Helping Proposition. A triangle ABC is given. Let AD, BE,
          CF be, it's altitudes and let H be, it's orthocenter. We draw
          the circumcircle (K1) of the cyclic quadrilateral BDHF and we
          denote as G, the intersection point of (K1), DE. Also we denote
          as L, the intersection point of BE, FG. We consider an arbitrary
          point between F, L and we denote as N, the intersection point of
          (K1), HP.
          If R, is the intersection point of BN, FG and Q, is the one
          of BP, EG, prove that the points Q, R, A, are collinear.

          7.1a Proof. We construct the line(e),through H and
          perpendicular to the line BE. We denote as F', R', L', the
          intersection points of HF, HR,(e)respectively, from the line BP.

          7.1b It is easy to prove that the segment line FG,is
          perpendicular to the BE ( from cyclic quadrilateral HDCE we
          have that <EHC = <EDC, etc ). So, the point P, is the
          orthocenter of the triangle RBH and then we have that the
          segment line BR', is perpendicular to HR. Hence the point R',
          lies on circle (K1) ( BH is diameter of (K1) ).

          7.1c The points Q, R, A, lie on the side lines BP, FP, BF
          respectively, of the triangle BFP. So, in order to these points
          are collinear, it is enough to prove that is true, the
          equality: [(RP):(RF)]·[(AF):(AB)]·[(QB):(QP)] = 1 (1)
          ( Menelaus theorem )

          Because of the segment lines FL, AE are parallel,
          => (AF):(AB) = (EL): (EB) (2)

          Also from the triangle BLP, by Menelaus theorem, we have that
          => [(EL):(EB)]·[(QB):(QP)]·[(GP):(GL)] = 1 (3)

          By (2),(3),the equality (1) becomes:(RP): (RF)=(GP):(GL) (4)

          From (4) => (RP–RF):(RF)=(GP–GL):(GL) => (FP):(RF)=(LP):(GL)

          => (FP):(RF) = (LP):(FL) (5) ( because of FL = GL )

          Hence it is enough to prove that is true the equality (5),
          instead of (1).

          7.1d We compare the bundles of lines B.LPFR and H.L'R'F'P.
          Their homologous lines, form congruent angles ( <LBP = <L'HR',
          <PBF = <R'HF', <FBR = <F'HP ).

          So, their cross-ratios are congruent. That is we have
          that ( L,P,F,R ) = ( L',R',F',P ) (6)

          Because of ( L',R',F',P ) = ( P,F',R',L' )
          => ( L,P,F,R ) = ( P,F',R',L' ) (7)

          Also because of PR //(e) we have that
          ( P,F',R',L' )=( P,F,R ) (8)

          From (7),(8),=> ( L,P,F,R ) = ( P,F,R ) (9)

          From (9) => [(FL):(FP)]:[(RL):(RP)] = (RP):(RF)

          => [(FL):(FP)]·[(RP):(RL)] = (RP):(RF)

          => (FL):(FP) = (RL):(RF) => (FP):(RF) = (FL):(RL) (10)

          7.1e From (5),(10),=>(LP):(FL)=(FL):(RL)=>(FL)²=(LP)·(RL) (11)

          From the orthogonal triangle BFH ( <BFH = 90º )
          =>(FL)²=(LB)·(LH) (12)

          From (11),(12), => (LP)·(RL) = (LB)·(LH) (13)

          Hence it is enough to prove that is true the equality (13),
          instead of (1).

          Really, this equality is true because of the similarity of the
          orthogonal triangles, PLH and BLR ( <PLH = 90º, <BLR = 90º ).

          We conclude now, that is also true the equality (1). Hence the
          points Q, R, A, are collinear and the proof of proposition
          (7.1), is completed.

          Sometimes ( many times for me ), our mind follows a complicated
          or a long way for the solution of a problem. So, I am waiting
          for your advices, for a simpler synthetic proof.

          I don't know if P ( the concurrency point of AD1, BE1, CF1 ), is
          a well known point in the plane of a triangle and I still not
          have studied, if it has any characteristic property.

          I will post here, next time, the proofs of Lemmas 1, 2 ( see
          mesage 13171 ) and also the ( easier ) proofs of propositions
          (9), (10),...,(15).
          I still not have any proof, for the proposition (8).

          Best regards.
          Kostas Vittas.
        • Kostas Vittas
          ... Dear all my friends. A friend of mine Kostas Helatiotis, said me that we don t need any helping proposition for the proof of (7), in message 13274. 7f.
          Message 4 of 4 , Jun 27, 2006
            --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
            wrote:
            > ...
            >
            >7f. As a helping proposition, we use the result that the points B',
            >A, S, are collinear ( we shall prove the proposition for this
            >result, below ). Hence, we consider the segments ...
            >
            > ...


            Dear all my friends.

            A friend of mine Kostas Helatiotis, said me that we don't need
            any helping proposition for the proof of (7), in message 13274.

            7f. From cyclic non-convex hexagon D'D''D1CEB, by Pascal's
            theorem, we have that the points B', A, S, are collinear. Hence, we
            consider the segments ...

            I apologize for my blindness and please, cancel all about the
            proof of the helping proposition (7.1), as unnecessary prattlings.

            Best regards.
            Kostas Vittas.
          Your message has been successfully submitted and would be delivered to recipients shortly.