In article <firstname.lastname@example.org>, email@example.com (Cecil Andrew Ellard) writes: > An old problem of plane geometry (Fermat's Problem) asks the > following: In a given acute triangle ABC, locate a point P whose > distances from A, B, and C have the smallest possible sum. What is > known about the related problem in which one drops the assumption of > acuteness, requires that the point P be on or inside the triangle ABC, > and instead asks for the largest possible product? To be precise, let > P(1)=(x(1),y(1)), P(2)=(x(2),y(2)), and P(3)=(x(3),y(3)) be three > distinct points in the plane which are not colinear. Therefore, the > convex hull of these three points (the smallest convex set containing > all three) is a triangular set. Call this triangular set T. Define a > function f from T to the real numbers by f(P) = the product of the > distances from P to the points P(1),P(2), and P(3). (So, for example, > f(P(1)) = f(P(2)) = f(P(3)) = 0.) Now here are the questions:(1) Does > the function f attain an absolute maximum on the set T?
T is a compact subset of R^2, so f, a continuous function from T to R, must attain its supremum. It need not do so at a unique point. For example, for an equilateral triangle, it seems fairly clear that there are three equal maxima at the mid-points of the sides.
> (2)If f does > attain an absolute maximum on T, can this maximum occur in the > interior of T?
A handwavy argument suggests no. My guess is that any maximum must lie on an edge of the triangle. I hope someone else can give a more satisfactory reply.
> (3) Is there a formula for the coordinates (using the > coordinates of P(1),P(2),and P(3)) of any point(s)where f attains an > absolute maximum?
As always, it depends on what you mean by formula, but for a reasonable sort of definition it looks as though the answer may be yes, but very complicated.
If the handwavy argument is correct then P must lie on an edge, and it seems we can find a formula for the best points on each edge.
To find the best point on edge AB, let's scale and rotate the triangle so that A=(0,0), B=(0,1), C=(c,d). We must then choose t in [0,1] to maximise f, or (equivalently and less uglily) to maximise
F = f^2 = t^2 (1-t)^2 ((t-c)^2 + d^2).
Clearly the max cannot be at t=0 or 1, so it must be at a point where dF/dt = 0. F is a sextic and dF/dt is a quintic, but factors into
dF/dt = 2 t (1-t) (a cubic),
and there's a formula for solving cubics.
If the handwavy argument, or rather its conclusion, is false, we'll have to solve F_x = F_y = 0, which I suspect is not possible in elementary functions.
> (4) Are any points where f attains an absolute > maximum constructible from P(1), P(2), and P(3) using straight-edge > and compass?
Seems unlikely unless the cubic in (3) has special properties I haven't spotted.
-- Robert Hill
Information Systems Services, University of Leeds, England