NavList:

I just got around to checking my email. Thank you Herbert for your explanation which finally made it through my thick head. I am posting your email and diagram for everyone else's benefit.

gl

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Gary,

I am sorry to make you wait so long. I am really not equipped to do geometry on the net. I made a sketch that I think will be helpful, but I never learned how to post it to the archive without mailing it to everybody on the list. Could you please upload this diagram for me?

Say we have three LOPs, a,b, c, intersecting in A, B, C, as per the attached diagram. V. shows through partial differentiation of the equations of condition that for the MPP the distances from the sides are as the sides. In the attached diagram I construct a line so that each point on it has the distances y and z from b and c, where y:z = b:c. (It so happens that this line is the symmedian of b and c, but this property is irrelevant for the construction.)

Draw a parallel to c at distance c, opposite to C
Draw a parallel to b at distance b, opposite to B
The parallels intersect at A'.
Draw a line through A' and A. This is the line with the above property.
Repeat with another pair and intersect the two results.

There is a practical problem with plotting this in that the point A' is often situated far outside the triangle. One remedy is to use distances b/2 and c/2. Another one is to draw both parallels on the opposite side from where I have drawn them for better clarity. I will show a better method shortly.

Herbert Prinz

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