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Re: [EMHL] A minimum problem (correction)

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  • Michael Keyton
    I see I misread the problem, that P was not on ell. I ll try again with a solutiondifferent from the ones given. Let A and B be on the same side of ell.
    Message 1 of 5 , Mar 5 7:52 PM
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      I see I misread the problem, that P was not on ell.
      I'll try again with a solutiondifferent from the ones given.
      Let A and B be on the same side of ell.
      Project A and B onto ell at A' and B' respectively.
      Construct equilateral triangle AA'A''so that A'' and B are on the same
      side of AA'. Let T be the intersection of AA'' and ell. If B is in the
      interior of triangle ATA', then B is the minimum point.
      if not, construct equilateral triangle BB'B'' so that B'' and A are on
      the same side of BB'. Let BB'' intersect ell at U. If A is in the
      interior of triangle BUB', then A is the fermat point (hence P=A).
      If not, then let P = thte intersection of AA'' and BB''. let P' be the
      projection of P on ell. Since A'APP' and B'BPP' are right trapezoids,
      and angle A'AP and B'BP are 60 deg, then PP' and BPP' are 120 deg. Which
      makes P the isogonal point of triangle ABP'.

      Michael Keyton

      Paul Yiu wrote:
      >
      > Dear Pat,
      >
      > Thank for pointing out the mistake in my statement in (2). Here
      > is a correction.
      >
      > [PY]: A and B are two points on the same side of a line ell. To
      > locate (construct if possible) a point P such that the sum of the
      > distances from P to A, B and ell is minimum.
      >
      > Solution:
      > Let H and K be the projections of A and B on the line ell. Construct
      > the equilateral triangle ABC such that C and ell are on opposite
      > sides of AB. Let the perpendicular from C to ell intersect the
      > circumcircle of ABC (again) at P and the line ell at Q.
      >
      > (1) If the vertex C is not between the perpendiculars AH and BK, then
      > the minimum occurs at A or B, whichever is closer to ell.
      >
      > (2) If C is between the lines AH and BK, and if P is on the other side
      > of ell as A and B, then the minimum occurs at Q. [This is the
      > case of the Fermat problem].
      >
      > (3) If C is between the lines AH and HK, and the circumcircle of ABC
      > does not intersect the line ell, then the minimum occurs at P. The
      > sum of the distances AP, BP and PQ is equal to CQ.
      >
      > I obtained this with the help of calculus, but the calculation is
      > exceedingly easy. I wonder if there is a pure synthetic reasoning.
      >
      > Best regards
      > Sincerely,
      > Paul
      >
      > At 03:40 PM 2/26/01 +0900, you wrote:
      > >Wow, it is so clear now.. I do have one minor question...
      > >
      > >In the part two of your solution, (the circumcircle intersects ell) it
      > >seems that if the point P is still on the same side of ell as ABC then it
      > >would still be a better solution than Q (this would be so that P is between
      > >C and Q). I only get Q better if P is on the opposite side of ell (Q is
      > >between C and P).
      > >
      > >If P is On the circle and thus APBC is a cyclic quadrilateral, the angle APB
      > >must always be 120 degrees, so anytime Q is on L and P is on the same side
      > >as AB, it would seem to be the Fermat point of AQB and therefore a better
      > >minimal solution.
      > >
      > >My apoligies if I simply misunderstand some of what you did.
      > >Pat Ballew,
      > >Misawa, Jp
      > >
      > >"Statistics means never having to say you're certain."
      > >
      > >
      > >Math Words & Other Words
      > >http://www.geocities.com/poetsoutback/etyindex.html
      > >
      > >The Mathboy's page http://www.geocities.com/poetsoutback/
      > >
      > >
      > >
      > >-----Original Message-----
      > >From: yiu@... [mailto:yiu@...]
      > >Sent: Monday, February 26, 2001 12:19 PM
      > >To: Hyacinthos@yahoogroups.com
      > >Subject: [EMHL] A minimum problem
      > >
      > >
      > >Dear friends,
      > >
      > >[PY]: A and B are two points on the same side of a line ell. To
      > >locate (construct if possible) a point P such that the sum of the
      > >distances from P to A, B and ell is minimum.
      > >
      > >Solution:
      > >Let H and K be the projections of A and B on the line ell. Construct
      > >the equilateral triangle ABC such that C and ell are on opposite
      > >sides of AB. Let the perpendicular from C to ell intersect the
      > >circumcircle of ABC (again) at P and the line ell at Q.
      > >
      > >(1) If the vertex C is not between the perpendiculars AH and BK, then
      > >the minimum occurs at A or B, whichever is closer to ell.
      > >
      > >(2) If C is between the lines AH and BK, and the circumcircle of ABC
      > >intersects the line ell, then the minimum occurs at Q. [This is the
      > >case of the Fermat problem].
      > >
      > >(3) If C is between the lines AH and HK, and the circumcircle of ABC
      > >does not intersect the line ell, then the minimum occurs at P. The
      > >sum of the distances AP, BP and PQ is equal to CQ.
      > >
      > >I obtained this with the help of calculus, but the calculation is
      > >exceedingly easy. I wonder if there is a pure synthetic reasoning.
      > >
      > >Best regards
      > >Sincerely,
      > >Paul
      > >
      > >
      > >
      > >
      > >
      > >Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
      > >
      > >
      > >
      > >
      > >
      > >Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
      > >
      > >
      > >
      >
      >
      >
      > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
    • Lambrou Michael
      ... Sorry if I miss something, but I don t understand the last step (and I don t mean the typo PP -- APP ): True enough, P is the isogonal point of ABP ,
      Message 2 of 5 , Mar 6 6:32 AM
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        On Mon, 5 Mar 2001, Michael Keyton wrote:

        > I see I misread the problem, that P was not on ell.
        > I'll try again with a solutiondifferent from the ones given.
        > Let A and B be on the same side of ell.
        > Project A and B onto ell at A' and B' respectively.
        > Construct equilateral triangle AA'A''so that A'' and B are on the same
        > side of AA'. Let T be the intersection of AA'' and ell. If B is in the
        > interior of triangle ATA', then B is the minimum point.
        > if not, construct equilateral triangle BB'B'' so that B'' and A are on
        > the same side of BB'. Let BB'' intersect ell at U. If A is in the
        > interior of triangle BUB', then A is the fermat point (hence P=A).
        > If not, then let P = thte intersection of AA'' and BB''. let P' be the
        > projection of P on ell. Since A'APP' and B'BPP' are right trapezoids,
        > and angle A'AP and B'BP are 60 deg, then PP' and BPP' are 120 deg. Which
        > makes P the isogonal point of triangle ABP'.
        >
        > Michael Keyton
        >

        Sorry if I miss something, but I don't understand the last step (and I
        don't mean the typo PP' --> APP'): True enough, P is the isogonal point of
        ABP', making AP+BP+PP' a minimum among points in ABP'. The question is,
        why is P' the right point on ell?
        For any P' on ell, findind the isogonal P of ABP' is easy, but that does
        not make P the minimum required by the original problem (not just the
        minimum of "something").
        Ahhh yes.
        In fact I don't think the constructed P is the right one, since it claims
        that the angle AP makes with ell is 30 degrees.This cannot be so because
        you can move B a little bit (without violating the condition about B
        being exterior to ATA'), but 30 remains 30.
        I think Paul's original solution using Ptolemy's theorem is from The
        Book.

        Michael.
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