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Bryan
guest
Mar-07-08, 00:40 AM (EST)

"Arcs of concentric circles"

 Not sure whether this belongs in High School or College math, but I suspect putting it on the college list would be flattering myself. Here is the general form of my question:For two concentric circles intersected by a single chord, what function (if any) relates their respective diameters and the arcs subtended by the chord on each of their circumferences?I am specifically interested in the case in which the arc on the smaller circle is 60 degrees. I am given to understand (and am trying to verify) that in this case, the arc of the larger circle is 144 degrees when the diameters of the two circles are related in mean/extreme ratio (i.e., the golden section). It may be that the fault is my poor compass/straightedge skills; but when I have tried this (sometimes following Euclid but usually approximating it by taking two Fibonacci numbers for the radii), I cannot confirm the proportions. But even if I could, I would want to understand the proof, which means (I think), understanding the general case.Thanks in advance for any help. ~~Bryan

Subject     Author     Message Date     ID
RE: Arcs of concentric circles alexb Mar-07-08 2
RE: Arcs of concentric circles Bryan Mar-07-08 3
RE: Arcs of concentric circles alexb Mar-07-08 4
RE: Arcs of concentric circles Bryan Mar-08-08 5
RE: Arcs of concentric circles Bryan Mar-08-08 6
RE: Arcs of concentric circles Bryan Mar-19-08 7
RE: Arcs of concentric circles alexb Mar-19-08 8

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alexb
Charter Member
2206 posts
Mar-07-08, 10:13 AM (EST)

2. "RE: Arcs of concentric circles"
In response to message #0

 >For two concentric circles intersected by a single chord, >what function (if any) relates their respective diameters >and the arcs subtended by the chord on each of their >circumferences? Let r and R be the radii of the two circles and h the distance from the center of the circles to the chord. There are isosceles triangles formed by the radii joining the center to the endpoints of the chords. These are divided by the altitude from the center to the chord into two right triangles. Let δ and Δ denote these right triangles for the small and the big circle. δ has angles 30°-60°-90°. Δ has angles 72°-18°-90°. From here,h = r·sin(60°),h = R·sin(18°). It follows thatR/r = sin(60°) / sin(18°).You surely know the value of sin(60°). The value of sin(18°) can be found at

Bryan
guest
Mar-07-08, 03:27 PM (EST)

3. "RE: Arcs of concentric circles"
In response to message #2

 Thank you so much. I followed your lesson and worked it out, just as you said. And lo! (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

alexb
Charter Member
2206 posts
Mar-07-08, 11:27 PM (EST)

4. "RE: Arcs of concentric circles"
In response to message #3

 (1/2) x 1/((sqrt5 - 1)/4) = phi!You should double check that because sin(60°) = √3 / 2.

Bryan
guest
Mar-08-08, 07:38 AM (EST)

5. "RE: Arcs of concentric circles"
In response to message #4

 you are right, I was hasty.rechecking. sigh.

Bryan
guest
Mar-08-08, 05:25 PM (EST)

6. "RE: Arcs of concentric circles"
In response to message #5

 So I left off (sqrt3) from my reckoning, carelessly, and oddly enough got the answer I thought I was looking for. Now I think I understand why I should have been wary. I ought not really to expect to find phi in the ratio R/r. Rather, it would be (if anywhere) in (R-r)/r. This is because it is the radius R which is (per hypothesis) divided by extreme & mean ratio.I'll try to apply the same principle you showed me and see if I get anywhere.~Bryan

Bryan
guest
Mar-19-08, 03:12 PM (EST)

7. "RE: Arcs of concentric circles"
In response to message #6

 Thought I would write again to let you know I followed through and verified that phi is *not* the ratio of R/r nor of R/(R-r) in the problem I originally presented. So the arc of 144 degrees is *not* related to the arc of 60 degrees by a chord on two concentric circles, when the respective radii are related to by the golden ratio.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.

alexb
Charter Member
2206 posts
Mar-19-08, 08:29 PM (EST)

8. "RE: Arcs of concentric circles"
In response to message #7

 >Which makes me feel a little >better about my steadiness with good old Euclidean tools. 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.