Does the Golden Spiral pass through the 4th vertex???

Hey Everyone:) I wasnt exactly sure which catergory this fitted by it involves complex analysis so i choose this one :) Okay my problem is in bold, the other questions i have solved but i just put them up there for background information just incase it was needed:

Using a golden triangle of base 12cm, draw a careful, accurate, neat logarithmic spiral.
Refer to Appendix 1 for the beginning instructions.

* Explain the relationship between Golden rectangle and Golden gnomon in creating this spiral.

* Measure some ratios of sides, how and where does appear in this construction?

* It is sometimes called the Equi Angular spiral. Why?

* Find the length of the spiral correct to the nearest millimetre. Show working.

* When you construct this Golden Spiral, does it pass through a 4th vertex of the pentagon formed from the
Golden Triangle? A reasonable proof must be given.

So we have a provide a reasonable proof that the arc 1 (the large one) does or does not pass through where the 4th vertex would be (the pentagon's point, which is just a Golden Gnomon added onto the side). It is obvious that is does not, but i need a proof :/ I have tried using trig and failed, due to not knowing the angles, i thought about using calculus but didnt think that would work either and im not very good at it yet so i;m not exactly sure about it, so im kinda stumped about this one. The only way i can think of is to measure it which obvious isn't a mathematical proof. Thanks heaps for any help :)

That's an interesting question. I'm referring to this picture and this one. I'd like a less algebraic approach than mine but I haven't found one.


Let the midpoint of the short side be at (0, 0) with the two vertices of that side at (-phi/2, 0) and (phi/2, 0) with the third vertex C above the x-axis. Let M1 be the midpoint of the line connecting (-phi/2, 0) and C; let M2 be the midpoint of the line connecting (-phi/2, 0) and (phi/2, 0). Cut off a golden triangle by drawing a line from (-phi/2, 0) to the opposite side, meeting it at point P. The distance from P to (phi/2, 0) is 1. Let R be the intersection of the lines M1 (phi/2, 0) and M2 P. This is the center of the logarithmic spiral, whose coordinates can be computed (I won't go through the computation).

Find the fourth point F of the pentagon. You can compute the distance between R and C, R and F, and R and (-phi/2, 0). You can also compute the angle between the lines RC and RF; and RF and R (-phi/2, 0). Since the spiral passes through C and (-phi/2, 0), you can compute its parameters. Finally you can compute the radius of the spiral at the angle the segment RF determines, which will be less than the radius RF.


If you find there's a neat geometric way, please describe it!

Yeah thanks heaps for answering it but due to time constraints i had to figure it out quickly and after four hours of trying i decided it was bed to to which the answer quickly jumped in my head. My way of solving it was that due to the base being 12, then the golden that came off it gave us the radious of the circle. So then the golden gnomon which has the arc situated within it height was able to be found to be 7.04cm, so therefore 2 golden gnomons the same size would give a height of 14.08 therefore the arc could not pass through the fourth vertex. Im sorry i wasnt able to upload a photo, my internet was being really slow that nite and really annoying :/
http://upload.wikimedia.org/wikipedi...spiral.svg.png

Thats what the spiral looked like anyways. Thanks heaps for your help and your time though, much appreciated :)