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Topic: Trigonometric area optimization
Replies: 13   Last Post: Dec 20, 2012 3:29 PM

 Messages: [ Previous | Next ]
 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: Trigonometric area optimization
Posted: Dec 16, 2012 11:55 AM

> ... I seem to come up with m = 3 (not
> exactly 3 maybe, but darned close). - 1/3 is the
> slope of the other line.

Hi Thor, bobmck and Peter,

In my posted reply of 11 Dec I treated the problem from First Principles, because that is the best way to understand a problem.

Since then bobmck has suggested a neat quick solution using a powerful, easily proved, general theorem on the minimum area of a triangle formed by a vertex angle (not necessarily a right angle) at a given point and a fixed line not thru the point. The triangle of minimum area is isosceles wrto the vertex.

Accepting the above theorem, we can proceed as follows:

It was shown that
OA = sqrt(1+m^2).10/(m+2) and
OB = sqrt(1+m^2).10/(2.m-1)

The required minimum triangle has OA = OB
.: m+2 = 2.m-1 so m=3 QED.

Regards, Peter Scales.

Date Subject Author
12/11/12 Peter Scales
12/14/12 bobmck
12/14/12 Peter Duveen