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simont

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Wed 2003-11-12 21:39
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[personal profile] simontThu 2003-11-13 01:05
but shortest distance and perpendicular are the same thing!

To an infinitely long line, yes, but in my previous wording there was the danger that someone might consider each edge of the polygon to be a finitely long line segment. The shortest distance to a line segment, if you're somewhere beyond one end of it, is nothing like the perpendicular distance to the containing line.

also... how is it irregular if convex and all the lengths are the same...

The angles are allowed to vary. For example, imagine a square of side 1, and an equilateral triangle of side 1. Join them together along one side, so that you get a pentagon looking a little like the end view of a house. This is convex, and all its side lengths are 1, but by no stretch of the imagination is it a regular pentagon.
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[identity profile] aiwendel.livejournal.comThu 2003-11-13 02:04
hmm ok i guess you COULD be beyond one end of it in a funny stretched ovaly polygon - and then you couldn't be perpendicular to it at all...

i've forgotten the rule now...
it can't work on irregular ones can it?
or was that the point?....

not woken up yet...
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[personal profile] simontThu 2003-11-13 02:15
It does work on irregular polygons, provided all their sides are the same length and they're convex. For example, it does work for the house-shaped pentagon I described above: for any point within the house shape, the sum of the perpendicular distances from all five sides is a constant.
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[identity profile] aiwendel.livejournal.comThu 2003-11-13 02:22
... *thinks* i was thinking of the dist between the point and the base... not the sides
oops.
still not 100% convinced... need to get a piece of paper!
is it the sum of ALL the sides then?
that would help :)
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