Date: Nov 14, 2001 12:46 AM
Author: Virgil
Subject: Re: equilateral triangle

In article <5.0.0.25.0.20011113093810.00a855e0@ltsp.com>,
 mary@krimmel.net (Mary Krimmel) wrote:

> Good, Virgil.
> 
> Now that you have found P, it looks as though we can show the equality 
> geometrically, by constructing the figure ACBP and dissecting it into eight 
> congruent 30-60-90 triangles.
> 
> I find your algebra hard to follow. In the second line of fifth paragraph, 
> isn't
> |AP| = |BP| = sqrt (x^2 + 1/4), rather than ...sqrt(x^2 + 1/2)?

Actually, I meant sqrt(x^2 + (1/2)^2), but yes.
> 
> But nevertheless you found a point P which gives the equality we looked for.


> 
> Nor do I understand why you have the proviso of your second paragraph. 
> Doesn't your construction show a sufficient point P whether or not ray CP 
> must intersect AB? Did you mean "may" rather than "must"? But even so, 
> nothing in the original problem prevented that possibility.
> 

In fact, there are such points P so long as the ray CP intersects the 
segment AB, but not otherwise. 

I said "may" because at that time I had only investigated the bisecting 
case.

The locus of such points P is a sort of arc from A to B bulging away 
from C. It does not appear to be a piece of a conic section, although it 
is close to parabolic, over  99% correlation.


> Thank you for your useful response. (Not my original posting, but the 
> problem did/does interest me.)
> 
> Mary Krimmel
> mary@krimmel.net
> 
> At 02:54 AM 11/13/2001 +0000, you wrote:
> >In article <5.0.0.25.0.20011112103226.00a73eb0@ltsp.com>,
> >  mary@krimmel.net (Mary Krimmel) wrote:
> >
> > > The second part looks easy. If P coincides with A then |AP| = 0 and |AP| +
> > > |BP| = |BP| = |CP|.
> > > Similarly we have equality if P coincides with B.
> > >
> > > I think otherwise we cannot have equality.
> >
> >Its possible, but not so easy as P = A and P = B.
> >
> >If ray CP must intersect side AB, then I think that there is a point P,
> >on the opposite side of AB from C for which we have equality.
> >
> >Specifically:
> >If ray CP bisects AB, and P is sqrt(3)/6 times the sidelength away from
> >AB on the side opposite C, then |AP|+|BP|=|CP|:
> >
> >Let the sides of the triangle be of length 1, and place the triangle in
> >a Cartesian coordinate system with A = (0,1/2), B = (0, -1/2), and
> >C = (-sqrt(3)/2,0), on the left of the y-axis.
> >
> >Then for P = (x,0), on the perpendicular bisector of segment AB,
> >|AP| = |BP| = sqrt(x^2 + 1/2), and |CP| = |x +sqrt(3)/2|,
> >so |AP|+|BP|=|CP| becomes 2*sqrt(x^2+1/2) = |x + sqrt(3)/2|.
                                        ^^^
                                        1/4, my error.

> >
> >Squaring and simplifying gives the  quadratic equation,
> >x^2 - sqrt(3)*x/3 + 12 = 0 = (x - sqrt(3)/6)^2.
> >
> >Thus x = sqrt(3)/6 and y = 0 gives |AP|+|BP|=|CP|, Q.E.F.
> 

Further analysis indicates that for any fixed y, with  |y| <= 1/2,
|AP|+|BP|-|CP| =
sqrt[x^2+(y-1/2)^2]+sqrt[x^2+(y+1/2)^2] - sqrt[(x + sqrt(3)/2)^2 + y^2]
has a unique minumum equalling zero at one positive value of x.

When |y| > 1, there is still appears to be a unique minumum, but it is 
strictly positive, so |AP|+|BP|-|CP| is not zero for |y| > 1, i.e., when 
CP does not pass through segment AB.