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Topic: Plane (Old) Geometry
Replies: 2   Last Post: Jul 6, 2000 12:30 PM

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Robert Hill

Posts: 529
Registered: 12/8/04
Re: Plane (Old) Geometry
Posted: Jul 5, 2000 1:52 PM
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In article <v31v2kik3v56@forum.mathforum.com>, emptyclass@yahoo.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






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