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Re: Trigonometric area optimization
Posted:
Dec 16, 2012 11:55 AM
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> ... 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.
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