```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.
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