For isosceles triangles T with apical angle inceeding pi/3 --- exasperated at having no contrapositive for "exceeding", I hereby coin one --- any point along the base serves as centre p. For collinear T (with area zero) any point on the line serves as p. WFL On 10/29/11, Dan Asimov <dasimov@earthlink.net> wrote:
That is almost completely correct, but there is still one case to consider.
--Dan
Richard wrote:
<< Is the following the answer?
If the triangle is equilateral, then all points inside the triangle serve (and all points outside if you assign signs to the distances). Otherwise the point is at the vertex with the largest angle. R.
I wrote: << Given an arbitrary triangle T in R^2, characterize p in R^2 such that the sum of distances
S(p) := d(p, L_1) + d(p, L_2) + d(p, L_3)
is minimized, where the L_j are the affine lines containing the sides of T.
"Things are seldom what they seem." --W.S. Gilbert
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