Golden Ratio with a Rusty Compass
In a 2005 article, K. Hofstetter, offered an elegant way of constructing the Golden Ratio with a rusty compass, i.e. a 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
The proof is straightforward.
First, BF = FM; so that the projection of F to AB (denoted K) is the midpoint of BM:
GM/GK = DM/KF.
It is convenient to assume AB = 4. Then
GM = 2/(√5 + 2) and GK = 5 - 2√5.
AG/BG = (√5 + 1)/2.
- K. Hofstetter, Division of a Segment in the Golden Section with Ruler and Rusty Compass, Forum Geometricorum, v 5 (2005), pp. 135-136