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:

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

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

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

  4. 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 $L.$

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

golden ratio by Quang Tuan Bui, 5-step construction

$P$ 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 $L:$

golden ratio by Quang Tuan Bui, 5-step construction,proof

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.

Golden Ratio

  1. Golden Ratio in Geometry
  2. Golden Ratio in an Irregular Pentagon
  3. Golden Ratio in a Irregular Pentagon II
  4. Inflection Points of Fourth Degree Polynomials
  5. Wythoff's Nim
  6. Inscribing a regular pentagon in a circle - and proving it
  7. Cosine of 36 degrees
  8. Continued Fractions
  9. Golden Window
  10. Golden Ratio and the Egyptian Triangle
  11. Golden Ratio by Compass Only
  12. Golden Ratio with a Rusty Compass
  13. From Equilateral Triangle and Square to Golden Ratio
  14. Golden Ratio and Midpoints
  15. Golden Section in Two Equilateral Triangles
  16. Golden Section in Two Equilateral Triangles, II
  17. Golden Ratio is Irrational
  18. Triangles with Sides in Geometric Progression
  19. Golden Ratio in Hexagon
  20. Golden Ratio in Equilateral Triangles
  21. Golden Ratio in Square
  22. Golden Ratio via van Obel's Theorem
  23. Golden Ratio in Circle - in Droves
  24. From 3 to Golden Ratio in Semicircle
  25. Another Golden Ratio in Semicircle
  26. Golden Ratio in Two Squares
  27. Golden Ratio in Two Equilateral Triangles
  28. Golden Ratio As a Mathematical Morsel
  29. Golden Ratio in Inscribed Equilateral Triangles
  30. Golden Ratio in a Rhombus
  31. Golden Ratio in Five Steps
  32. Between a Cross and a Square
  33. Four Golden Circles
  34. Golden Ratio in Mixtilinear Circles
  35. Golden Ratio in Isosceles Right Triangle, Square, and Semicircle

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