# Ulam's Spiral, Oppermann's diagonals and the Golden Ratio

With the help of our friends over here:
http://mathematica.stackexchange.com/questions/6219/ulams-spiral-with-oppermans-diagonals-quarter-squares
We created Ulam's Spiral with Oppermann's diagonals
We noticed an East-West symmetry with the starting points of the SW and NE diagonals and a North-South symmetry with the starting points of the NW and SE diagonals. We drilled down to the center of the spiral and added 4 lines to show this relationship

Assuming each regular edge is 1 unit, we can calculate the distance each diagonal's starting point is from our new center vertex, add them together, and then calculate the average. We got a surprising result.

$$\frac{1}{4} \left(\sqrt{\frac{5}{4}}+\sqrt{\frac{5}{4}}+\frac{1}{2}+\frac{1}{2}\right)=\frac{\phi }{2}$$

where $\phi$ is the Golden Ratio

We know that there exists a Golden Spiral, but this result is unexpected.

Can anyone explain this?

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I think this has little to do with the spiral and everything to do with the right triangle drawn in the center of it. The question basically reduces to what the $1$-$2$-$\sqrt{5}$ triangle has to do with the original geometric construction of the golden ratio. – anon May 31 '12 at 6:23

It the Law Of Small Numbers: "There aren't enough small numbers to meet the many demands made of them." (credit: Richard Guy). A problem with an integer lattice and simple diagonal lines is going to involve square roots of small numbers, and there are only a few small numbers around, so whatever the answer turns out to be it's going to be something well known, like 2, or $\sqrt2$, or the golden ratio.