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Topic ClosedPicture 34

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John Connor View Drop Down
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Direct Link To This Post Posted: 16.Oct.2013 at 16:17
Mr. Cooper's explanation works for me.  Thanks for the detailed explanation.
"Humans have a strength that cannot be measured. This is John Connor. If you are reading this, you are the resistance."

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GoplerGop View Drop Down
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Direct Link To This Post Posted: 17.Oct.2013 at 17:22
hi thnx John and Kent

I borrow Kent's concept into problem 34, and get it right with accident, but still not very clear explanation for how i got this.

How would my both circles set their center on the green line? Is there another way to explain it in math point of view or other points of view.

thx guys, love this group discussion, love it so much


oh the file is in below thx





Edited by GoplerGop - 17.Oct.2013 at 17:27
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GoplerGop View Drop Down
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Direct Link To This Post Posted: 17.Oct.2013 at 17:26


Ooops this is the right file, thx guys


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Kent Cooper View Drop Down
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Direct Link To This Post Posted: 18.Oct.2013 at 15:45
>>
How would my both circles set their center on the green line? Is there another way to explain it in math point of view or other points of view.
>>
 
It's because the two point locations specified in the Circle command's 3Point option are equidistant from that Line and directly opposite each other, because they are on hexagons Mirrored across it -- the centers of the Circles cannot therefore be anywhere but on that Line.  That's a way of laying it out that didn't occur to me, probably because I think of the 3Point option as requiring actual point locations, and I forget that you can use things like Tangent object-snap for them.
 
I notice that the drawing I attached before doesn't show the full series of green Points as I described [I had uploaded it first without them, and thought I had replaced the upload, but apparently that didn't work].  I have attached a corrected version, showing the full series of both green and red Points.  To do it purely mathematically, I think you would need to define those paths (which I'm now pretty sure are parabolic, though I'm not positive) and an edge of the hexagon with equations, and solve for the two pairs of simultaneous equations.
 
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