simont | (no subject)

(no subject)

Shinier still. drswirly has found a much better proof of my geometric theorem than mine, and in particular his proof demonstrates that the polygon doesn't need to be regular – it only has to be convex and have all sides the same length. Very pretty.

(Someone pointed out that I wrote ‘shortest distance’ where I meant ‘perpendicular distance’, as well. Each edge of the polygon should be considered to be extended as far as necessary.)

I should shut up about maths, really. Particularly since I've been doing maths at work for a few weeks and it's been quite stressful, and to my immense relief it all started working properly today so I can take a break and do something less taxing, so quite why I'm now wibbling on about maths in this diary for fun is beyond me.

Anyway. This evening is supposed to contain sofa therapy, so I shall return to the sofa.

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but shortest distance and perpendicular are the same thing!
also... how is it irregular if convex and all the lengths are the same... i'm presuming it can't float in and out at random, else i imagine it wouldn't work... whereas anything based on a cirle would... oval wounldn't all be the same, though with any shape as you get nearer one side you get further from the other etc.
ok i should be asleep now!
night night
xxxxxxxxxxxxxxx

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.

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...

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.

... *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 :)

quite why I'm now wibbling on about maths in this diary for fun is beyond me.

but maths is fun :-)

-m-

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