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Re: [EMHL] A simple construction

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  • xpolakis@otenet.gr
    Dear Paul, ... Just to small comments: Since the circles passing through A and tangent to BC at B,C, resp., have another intersection A , means that the same
    Message 1 of 3 , Apr 1, 2001
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      Dear Paul,

      You wrote:
      >Given triangle ABC, it is easy to construct the two circles,
      >each passing through A, and tangent to BC at one of the vertices B
      >and C.
      >
      >How does one construct the circle tangent to BC, and to each of these
      >two circles externally?
      >
      >There is of course a construction by inversion. I have found a very
      >simple construction of this circle, and verified the correctness by
      >calculations. I wonder if the following construction is known and if
      >one can find a synthetic proof:
      >
      >Let the bisector of angle A intersect BC at P. Construct the
      >perpendicular to BC at P to intersect the median AD at Q.
      >The cirlce with PQ as diameter is tangent BC and to each of the two
      >circles above.

      Just to small comments:

      Since the circles passing through A and tangent to BC at B,C, resp.,
      have another intersection A', means that the same construction should
      apply for the triangle A'BC.

      Since there are two solutions of the problem (ie the circle in question
      touches BC between B, C or at a point on its extension), means that there
      should be a similar construction for the second circle.
      (in this case the bisector should be the bisector of the external angle A).


      Antreas
    • xpolakis@otenet.gr
      Dear Paul, ... Here is my proof: Assume that AC AB or b c A / / / / / K / M | / | / Q
      Message 2 of 3 , Apr 1, 2001
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        Dear Paul,

        You wrote:
        >Given triangle ABC, it is easy to construct the two circles,
        >each passing through A, and tangent to BC at one of the vertices B
        >and C.
        >
        >How does one construct the circle tangent to BC, and to each of these
        >two circles externally?
        >
        >There is of course a construction by inversion. I have found a very
        >simple construction of this circle, and verified the correctness by
        >calculations. I wonder if the following construction is known and if
        >one can find a synthetic proof:
        >
        >Let the bisector of angle A intersect BC at P. Construct the
        >perpendicular to BC at P to intersect the median AD at Q.
        >The cirlce with PQ as diameter is tangent BC and to each of the two
        >circles above.

        Here is my proof:

        Assume that AC > AB or b > c

        A
        /\
        / \
        / \
        / \
        / \ K
        / M |
        / \ |
        / Q \ |
        / \ |
        / N-------------L
        / | \|
        B-----H---P--D---------|C

        AP: int. bisector of ang(BAC) = A
        AD: a-median (D = midpoint of BC)
        AH: a-altitude

        M: midpoint of AC

        Q = AD /\ perp. to BC at P

        N = Midpoint of QP = center of the circle with diameter QP

        K = (perp. to AC at M) /\ (perp. to BC at C) : it is the center of the circle
        passing through A and tangent to BC at C (KC = radius of this circle)

        L= orth. projection of N on KC.

        PQ PD AH * PD
        Tr.(HAD) ~ Tr.(PQD) ==> ---- = ---- ==> PQ = -------
        AH DH DH

        We have:

        PD = PC - DC, PC = ab/(b+c), DC = a/2, DH = (b^2 - c^2)/2a

        So, we get: PQ = AH * a^2 / (b+c)^2.

        To prove that the circle (N, PQ/2) is tangent externally to circle (K, KC)
        is enough to prove that: KN = radius of the one circle + rad. of the other
        circle
        that is: KN = KC + NP or

        KN^2 + KC^2 + NP^2 + 2KC*NP (1)

        We have: KN^2 = KL^2 + LN^2 = (KC-LC)^2 + LN^2 = (KC - NP)^2 + PC^2 =

        KN^2 = KC^2 + NP^2 - 2KC*NP + PC^2 (2)

        From (1) and (2) we get that it is enough to prove that PC^2 = 4KC*NP (3).

        Is this last equality true? Let's see:

        We have:

        PC^2 = a^2b^2 / (b+c)^2, NP = PQ/2 = (AH*a^2)/2(b+c)^2

        KC MC KC b
        Tr. (MKC) ~ Tr.(HCA) ==> ---- = ---- or ---- = ---- ==> KC = b^2 / 2AH
        AC AH b 2AH


        So, the (3) becomes: a^2b^2 / (b+c)^2 = 4 * (b^2 / 2AH) * [(AH*a^2)/2(b+c)^2]

        ==> 1 = 1. True!

        Similarly we prove that the circle (N, PQ/2) touches externally the circle
        passing through A and tangent to BC at B.

        Antreas
      • xpolakis@otenet.gr
        ... [I changed the notation of the points. Similarly we define the points B , C ; B , C ; Pb, Pc] 1. Let Ma, Mb, Mc be the centers of Paul s circles (ie
        Message 3 of 3 , Apr 2, 2001
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          [PY]:
          >>Given triangle ABC, it is easy to construct the two circles,
          >>each passing through A, and tangent to BC at one of the vertices B
          >>and C.
          >>
          >>How does one construct the circle tangent to BC, and to each of these
          >>two circles externally?
          >>
          >>There is of course a construction by inversion. I have found a very
          >>simple construction of this circle, and verified the correctness by
          >>calculations. I wonder if the following construction is known and if
          >>one can find a synthetic proof:
          >>
          >>Let the bisector of angle A intersect BC at A'. Construct the
          >>perpendicular to BC at A' to intersect the median AA" at Pa.
          >>The cirlce with A'Pa as diameter is tangent BC and to each of the two
          >>circles above.
          [I changed the notation of the points. Similarly we define the points
          B', C'; B", C"; Pb, Pc]

          1. Let Ma, Mb, Mc be the centers of Paul's circles (ie midpoints
          of A'Pa, B'Pb, C'Pc).
          Are the lines AMa, BMb, CMc concurrent?

          2. Let P'a, P'b, P'c be the reflections of Pa, Pb, Pc in BC, CA, AB, resp.
          Are the lines AP'a, BP'b, CP'c concurrent?

          3. Let M'a, M'b, M'c be the reflections of Ma, Mb, Mc on BC, CA, AB, resp.
          Are the lines AM'a, BM'b, CM'c concurrent?

          Antreas


          PS: A typo in my proof:

          > KN^2 + KC^2 + NP^2 + 2KC*NP (1)
          ^

          KN^2 = KC^2 + NP^2 + 2KC*NP (1)
          aph
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