The perpendicular bisectors of the sides of a triangle meet in a single point.

[ Editor: posed by N.D. Aparicio <nda@eng.cam.ac.uk>. ]

If x is the distance along the side 1-2, and L is the length of the
side, then I think the most general symmetric (same solution if you flip
the triangle) solution is

x = L/2 + [ f(w1) - f(w2) ] / L

where f(w1,w,l) is an arbitrary function depending on w1,w,l where w is
a symmetric function of the wi's and l is a symmetric function of the
L's.  You can safely ignore w and l and consider f(w1).  A good f might
be

f(w1) = L0^2 * arctan(w1) / PI

where L0 is the length of the smallest side.  This function keeps the
intersection point inside the triangle.

-- David Wayne Ring <dwr2560@tamuts.tamu.edu>

[ Editor: Aparicio confirmed this as the correct answer and asked how to
  prove it.  Ring replied: ]

...and furthermore, that _any_ such symmetric formula is of this form.

I will leave your question open for now, because it is a good puzzle.
As a hint: consider changing one weight.  The line from the opposite
side will not change position.  That gives a condition on the changes in
the two other lines.  You will find that the most general solution is
simpler if you demand that the rule be symmetric between the vertices.

-- David Wayne Ring

Let wi be the weight of the vertex i and di(x,y) the distance between
an arbitrary point (x,y) and the vertex i.  Let us consider the function

fi(x,y) = di(x,y)*di(x,y) - h(wi)

where h(wi) is an arbitrary function (although it is convenient to
limit the set where this function can be chosen such as to guarantee
that all orthogonal lines will be drawn within the corresponding side
of the triangle).

There are three functions fi(x,y), one for each vertex.  Now, let
us define that region of the plane where f1(x,y) gives a value
which is less than those given by f2(x,y) and f3(x,y), and similarly
with f2(x,y) and f3(x,y).  It is clear that the whole plane must
be divided into three parts:   That where all points make f1(x,y) less
than f2(x,y) and f3(x,y), the other where all points make f2(x,y)
less than the other two and the third one where f3(x,y) gives the
least value, there can not be a fourth region, therefore, the
boundaries of these three regions (which are lines) must intersect
each other at one single point.

A boundary, for example between regions 1 and 2, will be defined
with the condition:

f1(x,y) = f2(x,y)   which leads to

d1(x,y)^2 - h(w1) = d2(x,y)^2 - h(w2)         (*)

The squares cancel, so this is actually the equation of a line,
furthermore, this line is parallel to a line defined by

d1(x,y)^2 = d2(x,y)^2

which is an orthogonal line to the segment 12 and drawn through its
middle point, therefore the boundary of regions 1 and 2 is a line
which is orthogonal to the segment 12 as well.

If d is the length of the segment 12, by substituting d2 = d - d1
into equation (*) we will get, after some arrangements:

d1/d = (1/2)*[1 + (h(w1) - h(w2))/d^2] [Editor: closing parenthesis after h(w2) missing in the original]

which is the same solution that Dave Ring from rec.puzzles and
Robert Israel from sci.math obtained (the factor 1/2 can be
introduced into the arbitrary function h).

-- Nairo Aparicio <nda@eng.cam.ac.uk>