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[–] SithEmpire 0 points 2 points (+2|-0) ago 

Pythagoras!

Basically you have a right-triangle with a base of 1/2 and a height of 1, so the diagonal is the square root of 1/4 + 1 = 5/4, or half the square root of 5.

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[–] assert_patrimony [S] 0 points 1 points (+1|-0) ago 

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[–] SithEmpire 0 points 1 points (+1|-0) ago 

Exactly like that; I would also note that having the rest of the outer square in the picture is probably just confusing things.

In any case, that diagonal will then be the golden ratio less 1/2 (or if you like, the diagonal plus the base is the golden ratio relative to the height).

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[–] Chommers 0 points 1 points (+1|-0) ago 

Sorry I don't want to make a picture but to me the simple way to think about this is as a linear relationship;

4 sides times your unit of 1 is four.

Now we want to know how long some unit is based on what we know about the container of that thing.

The midpoint of the starting line is 1/2 of one of our sides and then endpoint is a further 1/2 along on the same axis. So we can take those 1/2 and 1/2 and add them to our sides for 5.

Now to describe the intersection using this same 'scale' we find the line intersects the square 2 vectors apart. (Counting the side of the square and the bottom of the triangle as segments.)

Using such a fundamental way to describe this is helpful for describing more complex shapes later.

If you want to get into a better way of describing things how about fractals?

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[–] assert_patrimony [S] 0 points 0 points (+0|-0) ago  (edited ago)

I'm sorry you lost me at adding our sides to make 5. I think I got it, though. Thank you.

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[–] Chommers 0 points 1 points (+1|-0) ago 

Yeah there is a video on phi that will teach you a lot more. It will show how this concept relates to fibonacci sequence etc.

I was trying to help 'visualize' the math since that is the only way I can remember most things. Good luck with your mid-term!

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[–] Chommers 0 points 1 points (+1|-0) ago 

I'm not a math person and I probably didn't explain it properly. However I looked at 'Phi' with a Pentagon and the concept of counting sides and segments (as vectors) seems to apply there as well. The only difference with Φ is counting complete sides as segments.

So on a rectangle I counted the sides and added the length of the two offsets to get 5, but in a Pentagon finding Φ I count the two sides between the points and then add the three sides between the points on the other side, and still maintain your relationship with the 2 vectors back along the first two sides to get the 2.

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[–] MrPaul 1 points 0 points (+1|-1) ago 

The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side.

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[–] Wrathfullyawakenedww 0 points 1 points (+1|-0) ago 

Thank you scarecrow... the wizard will present you with a diploma... you got a brain!!!!