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Orthocenter as perspector

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  • Juan Carlos Salazar
    Dear friends: Let ABC be triangle. The tangency points of its excircles (Ia),(Ib), (Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K), (H,I,L)
    Message 1 of 5 , Oct 3, 2004
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      Dear friends:
      Let ABC be triangle. The tangency points of its excircles (Ia),(Ib),
      (Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),
      (H,I,L) respectively.Now we have:
      1)EF cut ID at X,GH cut EF at Y,GH cut DI at Z, so XYZ and ABC are
      perspective with perspector the orthocenter of ABC.
      2)XJ,YK,ZL concur at P. I thing that P isn't ETC.
      Can somebody find the barycentrics of P?
      Greetings
      Juan Carlos
    • Eric Danneels
      Dear Juan Carlos, you wrote ... (Ib), ... The first barycentric coordinate of P is (b+c)(a^2(3a^2-2b^2-2c^2)-(bb-cc)^2)/(b+c-a) This point is not in the ETC
      Message 2 of 5 , Oct 3, 2004
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        Dear Juan Carlos,

        you wrote

        > Let ABC be triangle. The tangency points of its excircles (Ia),
        (Ib),
        > (Ic) with (AB,AC,BC),(AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),
        > (H,I,L) respectively.Now we have:
        > 1)EF cut ID at X,GH cut EF at Y,GH cut DI at Z, so XYZ and ABC are
        > perspective with perspector the orthocenter of ABC.
        > 2)XJ,YK,ZL concur at P. I thing that P isn't ETC.
        > Can somebody find the barycentrics of P?

        The first barycentric coordinate of P is

        (b+c)(a^2(3a^2-2b^2-2c^2)-(bb-cc)^2)/(b+c-a)

        This point is not in the ETC (until 2445)

        If you replace J, K and L by the touchpoints of the incircle
        the perspector becomes X(1439) with first barycentric coordinate
        a(b+c)(b^2+c^2-a^2)/(b+c-a)^2

        Greetings from Bruges

        Eric
      • Darij Grinberg
        Dear Juan Carlos and Eric, Very nice results! In Hyacinthos message #10588, Juan Carlos Salazar ... Indeed, if Ia, Ib, Ic are the excenters of triangle ABC,
        Message 3 of 5 , Oct 3, 2004
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          Dear Juan Carlos and Eric,

          Very nice results!

          In Hyacinthos message #10588, Juan Carlos Salazar
          wrote:

          >> Let ABC be triangle. The tangency points of
          >> its excircles (Ia),(Ib),(Ic) with (AB,AC,BC),
          >> (AB,BC,AC),(AC,BC,AB) are (D,E,J),(G,F,K),
          >> (H,I,L) respectively.Now we have:
          >> 1)EF cut ID at X,GH cut EF at Y,GH cut DI at
          >> Z, so XYZ and ABC are perspective with
          >> perspector the orthocenter of ABC.
          >> 2)XJ,YK,ZL concur at P.

          Indeed, if Ia, Ib, Ic are the excenters of
          triangle ABC, then the lines YZ, ZX, XY are the
          reflections of the lines KL, LJ and JK in the
          lines IbIc, IcIa, IaIb, respectively. (Why?
          Well, since the two tangents from a point to a
          circle have equal length, we have BI = BL, and
          since the line IcIa bisects the angle IBL, it
          follows that the point I is the reflection of
          the point L in the line IcIa. Similarly, the
          point D is the reflection of the point J in the
          line IcIa, and hence, the line ZX, which is the
          same as the line ID, is the reflection of the
          line LJ in the line IcIa. Similarly for the
          lines XY and YZ.)

          Now, this implies that the lines YZ, KL and
          IbIc concur at one point X1, the lines ZX, LJ
          and IcIa concur at one point Y1, and the lines
          XY, JK and IaIb concur at one point Z1. These
          three points X1, Y1 and Z1 are collinear after
          the Desargues theorem, since the triangles
          IaIbIc and JKL are perspective (the perspector
          is X(40)). Hence, the Desargues theorem again
          yields that the triangles IaIbIc and XYZ are
          perspective, and that the triangles XYZ and
          JKL are perspective. This proves your property
          2).

          As for property 1), it is proven in the note
          "Synthetic proof of Paul Yiu's excircles
          theorem" on my website

          http://de.geocities.com/darij_grinberg

          (you probably know this, Juan Carlos, but I
          say this for others who are interested).

          In Hyacinthos message #10589, Eric Danneels
          wrote:

          >> If you replace J, K and L by the
          >> touchpoints of the incircle the perspector
          >> becomes X(1439)

          Indeed, if J', K' and L' are the points where
          the incircle of triangle ABC touches the sides
          BC, CA, AB, then the perspectivity of
          triangles XYZ and J'K'L' can be proven
          synthetically as well: A simple application of
          the Menelaos theorem shows that the lines YZ,
          K'L' and BC concur at one point X2, the lines
          ZX, L'J' and CA concur at one point Y2, and
          the lines XY, J'K' and AB concur at one point
          Z2. Now, these three points X2, Y2, Z2 are
          collinear, since we can apply the Desargues
          theorem to the two perspective triangles ABC
          and J'K'L' (they are perspective at X(7)).
          (Or we could also apply it to the triangles
          ABC and XYZ, perspective to each other at
          X(4).) And by Desargues again, we obtain the
          perspectivity of triangles XYZ and J'K'L'.

          Sincerely,
          Darij Grinberg
        • Juan Carlos Salazar
          Dear Eric, Darij and Paco García: Thank you very much for your remarks. Futher, if we consider:IL cut KF at U,HL cut JE at V,DJ cut KG at W. Also UVW and ABC
          Message 4 of 5 , Oct 3, 2004
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            Dear Eric, Darij and Paco García:
            Thank you very much for your remarks.
            Futher, if we consider:IL cut KF at U,HL cut JE at V,DJ cut KG at W.
            Also UVW and ABC are perspective with orthocenter of ABC as
            perspector
            Now UVW and JKL are perspective with perspector Q.
            Q is the Spieker center of ABC?
            Greetings
            Juan Carlos
          • Darij Grinberg
            Dear Juan Carlos, ... Yes, I have checked this by calculation. Unfortunately, I don t have a synthetic proof right now. Interesting result! Sincerely, Darij
            Message 5 of 5 , Oct 4, 2004
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              Dear Juan Carlos,

              In Hyacinthos message #10592, you wrote:

              >> IL cut KF at U,HL cut JE at V,DJ cut KG
              >> at W.
              >> Also UVW and ABC are perspective with
              >> orthocenter of ABC as perspector
              >> Now UVW and JKL are perspective with
              >> perspector Q.
              >> Q is the Spieker center of ABC?

              Yes, I have checked this by calculation.
              Unfortunately, I don't have a synthetic proof
              right now. Interesting result!

              Sincerely,
              Darij Grinberg
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