(1/2) x 1/((sqrt5 - 1)/4) = phi!

The piece I was missing was the line segment you called h, the distance from the center of the circles to the chord.

Thank you for explaining by appealing to the general trig functions. Although this was where I bottomed out in high school, when I worked at it, it was easy to see why the result obtains: since you start out with the general (sin theta) and then plug in a particular value after that, I can see that the result of phi obtains only when the two angles in question are 60 and 18 degrees. This was exactly the understanding I needed to get.

Very helpful. Thank you.
~~Bryan

You should double check that because sin(60°) = √3 / 2.

I'll try to apply the same principle you showed me and see if I get anywhere.

~Bryan

However, I decided to check and see what larger arc *was*, if i stipulated the 60-degree arc on the smaller circle and the radii being in that ratio.

Let there be two circles with a common center O. A chord subtends an arc of 60 degrees on the smaller circle, and a to-be-determined arc on the larger. Let h be the distance from O to this chord; it makes a right angle with the chord at point H, slightly inside the circumference of the smaller circle.

The radius r of the smaller circle is 1; The distance from the circumference of the smaller circle to the larger is phi. So the radius R of the larger circle = 1+phi.

There are straight lines connecting the center O to the intersection of the chord to the circumferences of the two circles. On the smaller circle the point of this intersection (call it A) is at a 60 degree angle. On the larger circle the point (call it B) is at an angle to be determined.

Since r=1, sin A= sin60= h/1 = h. So h= sin60 = ((sqrt3)/2)

sin B = ((Sqrt3)/2) / (1 + phi)
= (sqrt3)/(2(1 + phi))

so B = arcsin (sqrt3)/(2(1 + phi))

= approx 19.31687 degrees

(I confess I turned to an arcsin calculator, and used a long decimal approximation of phi). Now I could easily calculate the remaining angle of the other angle in triangle HOB. The I subtracted 30 degrees to get angle AOB.

Multiply AOB by 2, and add 60 degrees, and voila!
(naturally, I could also have just doubled HOB)

It isn't 144, but much closer to the 141 or so which I kept getting when I had tried with compass and straightedge and measuring with a protractor. Which makes me feel a little better about my steadiness with good old Euclidean tools.

Thanks again for your help, without which I would not have got started.

Your satisfaction is well deserved. It would have been nice to get the result you wanted but I could not see a way to get away with sqrt(3).

>Thanks again for your help, without which I would not have
>got started.

It was a pleasure helping in your case.