# 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 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.

The proof is straightforward. First, BF = FM; so that the projection of F to AB (denoted K) is the midpoint of BM: KM = BK. Right triangles GMD and GKF are similar which gives a proportion:

GM/GK = DM/KF.

It is convenient to assume AB = 4. Then FM = BF = 4, AM = BM = 2, KM = BK = 1. Solving right triangles we obtain KF = 15 and DM = 12 such that GM/GK = 2/5. Since GM + GK = KM = 1, we derive

GM = 2/(5 + 2) and GK = 5 - 25.

Further, AG = AM + GM = 2(5 - 1) and BG = GK + BK = 6 - 25. Now by direct verification,

AG/BG = (5 + 1)/2.

### References

1. K. Hofstetter, Division of a Segment in the Golden Section with Ruler and Rusty Compass, Forum Geometricorum, v 5 (2005), pp. 135-136 ### Fibonacci Numbers 