Let L and M be the midpoints of the sides AB and AC of an equilateral triangle ABC. Let X, Y be the intersections of LM extended with the circumcircle of ΔABC. Then LM / MY = φ.
Indeed, if 2a is the side length of ΔABC, then AM = MC = LM = a and XL = MY = b. By the Intersecting Chords Theorem,
MX·MY = AM·MC.
In other words,
(a + b)·b = a·a.
Denoting a/b = x, we see that
1 + x = x2,
which is (2), so that indeed x = φ. A derivation based on the presence of similar triangles was posted by Jan van de Craats as a solution to Coxeter's problem and included by R. Nelsen in his collection Proofs Without Words II.
(Linda Fahlberg-Stojanovska made a camcast that follows Odom's construction by paper folding a circle. The resulting pentagon is very nearly regular, but not quite.)
In the above mentioned article, K. Hofstetter, offered another elegant way of obtaining the Golden Ratio.
It will be convenient to denote S(R) the circle with center S through point R. For the construction, let A and B be two points. Circles A(B) and B(A) intersect in C and D and cross the line AB in points E and F. Circles E(B) and F(A) intersect in X and Y, as in the diagram. Because of the symmetry, points X, D, C, Y are collinear. The fact is CX / CD = φ. (The proof has been placed on a separate page.)
In a subsequent paper, Hofstetter gave a 5-step algorithm for dividing a segment in golden ratio:
Draw A(B) and B(A) and let C and D be their points of intersection. Draw C(A) and let it intersect A(B) in E and CD in F. Draw E(F). This intersects the line AB in points G and G' such that AB:AG = φ and AG':AB = φ.
For a proof, suppose AB has unit length. Then CD = √3 and EG = EF = √2. Let H be the orthogonal projection of E on the line AB. Since HA = 1/2, and HG2 = EG2 - EH2 = 2 - 3/4 = 5/4, we have AG = HG - HA = (√5 - 1)/2. This shows that G divides AB in the golden section.
In the third paper in the series, Hofstetter mentions having been alerted to the fact that the above proof had been discovered previously by E. Lemoine (1902) and L. Reusch (1904). The essence of the paper is an additional 5-step division of a segment into the golden section.
The construction is this. For a given segment AB of length 1, form circles A(B) and B(A). Let C and D be the intersections of A(B) and B(A). Extend AB beyond A to the intersection E with A(B). Draw E(B) and let F be the intersection of E(B) and B(A) farther from D. DF intersects AB in G. AG:BG = φ.
To see why this is so, let I be the intersection of CD and AB and H the foot of perpendicular from F to AB.
IG/GH = DI/FH = (√3)/2 / (√15)/4 = 2/√5.
It follows that IG = 2·IH/(√5 + 2) = (√5 - 2)/2, and AH = 1/2 + IG = (√5 - 1)/2. This shows that G divides AB in the golden section.
In a 2005 paper, Hofstetter offers a similar construction with a rusty compass whose opening can be set only once.
Draw A(B) and B(A) and find C and D at their intersection. Let M be the midpoint of AB found at the intersection of AB and CD. Construct C(M, AB), a circle with center M and radius AB. Let it intersect B(A) in F and another point, F being the farthest from D. Define G as the intersection of AB and DG. G then is the sought point. (For details of the proof check a separate page
The golden ratio has been sighted in a trapezoid [Tong]. In the diagram, the bottom base PQ has length b and the top base RS = a < b. The line MN parallel to the bases is of length
√(a² + b²)/2. This quantity that is known as the quadratic mean (or the root-mean-square) fits between a and b as any other mean would.
From the similarity of triangles MSV and PSW,
if b = 3a. The construction of the golden ratio based on this is depicted below
In the construction, BC = 3·AD, CE = AD, CE 
BC, BF = EF = FH, FH 
BE, BI = BH, IJ||AB, and GJ||BC. AG/BG = φ.
