## Re: [EMHL] A minimum problem (correction)

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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.
Message 1 of 5 , Mar 5, 2001
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/
• ... 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, 2001
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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