- 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/ - On Mon, 5 Mar 2001, Michael Keyton wrote:

> I see I misread the problem, that P was not on ell.

Sorry if I miss something, but I don't understand the last step (and I

> 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

>

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.