# Golden Ratio in Five Steps

Bùi Quang Tuån has devised a 5-step construction of the Golden Ratio (earlier 5-step constructions are due to K. Hofstetter.)

Here are Bùi Quang Tuån's five steps:

Draw line $L$ and choose point $O$ on that line.

Draw circle $(O)$ of a random radius and mark the points of intersection of $(O)$ with $L,$ say $A$ and $B.$

Draw circle $B(A)$ centered at $B$ and passing through $A.$ Let $C$ be the second intersection of $B(A)$ with $L.$

Draw circle $C(O)$ centered at $C$ and passing through $O.$ Let it intersect $B(A)$ at $N$ below $L$ and $(O)$ at $M$ above $L.$

Join $MN$ and let $P$ be the intersection of $MN$ and $(O).$

$P$ divides $MN$ in the golden ratio.

### Proof

The proof is much less elegant than the construction itself. First of all, let's choose the Cartesian coordinates with $O$ as the origin and $OA$ as $x$-axis.

$M$ is one of the intersections of two circles $x^{2}+y^{2}=1$ and $(x+3)^{2}+y^{2}=3^{2}.$ From these, $\displaystyle x=\frac{1}{6}$ and $\displaystyle y=\frac{\sqrt{35}}{6}.$ We can identify $M$ with $\displaystyle \bigg(-\frac{1}{6},\frac{\sqrt{35}}{6}\bigg).$

For $N,$ we similarly have two equations: $(x+1)^{2}+y^{2}=2^{2}$ and $(x+3)^{2}+y^{2}=3^{2}.$ Solving these gives $\displaystyle N=\bigg(-\frac{3}{4},-\frac{3}{4}\sqrt{7}\bigg).$

We can now find that $\displaystyle MN^{2}=\frac{7}{4}(3+\sqrt{5}).$ To establish the validity of the construction we also need to compute $MP.$ Bùi Quang Tuån suggested that it is easier to compute $DN$ where $D$ is the second intersection of $C(O)$ with $L:$

This is because in isosceles triangles $MOP$ and $DCN,$

$\angle CDN=\angle ODN=\angle OMN=\angle OMP.$

And, since the radii of the two circles are in the $1:3$ ratio, $DN=3\cdot MP.$ With our choice of the coordinates, $D=(-6,0)$ so that $\displaystyle DN^{2}=\frac{63}{2},$ making $MP^{2}=\displaystyle\frac{7}{2}.$

Finally, $\displaystyle\frac{MN^{2}}{MP^{2}}=\frac{3+\sqrt{5}}{2}=1+\phi$ which shows that $\displaystyle\frac{MN}{MP}=\phi,$ as required.

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