# 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.

Further,

AG/BG = (√5 + 1)/2.

### References

- K. Hofstetter,
__Division of a Segment in the Golden Section with Ruler and Rusty Compass__,*Forum Geometricorum*, v 5 (2005), pp. 135-136

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