Four Golden Circles

Bùi Quang Tuån has posted on the CutTheKnotMath facebook page a configuration of four circles and a parallelogram rich in interrelations between the parts among which are several sightings of the Golden Ratio.

Main statement

Each of the four congruent circles $(A),$ $(B),$ $(C),$ $(D)$ touches other two and passes through the center of the remaining third, as shown:

four golden circles by Quang Tuan Bui

In the parallelogram $EHFG,$

  1. The sides are in Golden Ratio: $FG/EG=\phi,$

  2. The diagonals are in Golden Ratio: $EF/GH=\phi,$

  3. Circle centers divide the sides in Golden Ratio, e.g., $AE/AG=BE/BH=\phi,$

  4. The two center lines $AB$ and $CD$ are divided in Golden Ratio by the diagonal $EF,$ e.g. $AI/BI=\phi,$

  5. Two lines $BF$ and $CE$ are divided in Golden Ratio by the diagonal $GH,$ e.g. $EK/CK=\phi,$

  6. In turn, lines $BF$ and $CE$ divide the diagonal $GH$ in Golden Ratio, e.g., $GL/HL=\phi.$

Note: The configuration has several additional properties worth of consideration. These are listed at the bottom of the page.


Without loss of generality, assume that the common radius of the four circles is $1.$ Let $N$ be the second intersection of $(A)$ and $(B);$ $M$ the midpoint of $AB.$

four golden circles by Quang Tuan Bui, step 1

Due to the symmetry of the configuration, $M$ is the intersection of $AB$ and $CE,$ $\Delta ABC$ is isosceles, with $AB=1,$ $AC=BC=2.$ By the Pythagorean theorem, $CM=\sqrt{15}/2.$ In equilateral $\Delta ABE,$ sides are equal to $1$ so that $EM = \sqrt{3}/2.$ By symmetry, also $MN=\sqrt{3}/2.$ Thus

$\displaystyle CE=\frac{\sqrt{15}}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}(\sqrt{5}+1)$

and $\displaystyle\frac{CE}{EN}=\frac{\sqrt{5}+1}{2}=\phi.$

Let now $I$ and $J$ be the intersections of $EF$ with $AB$ and $CD,$ respectively. By symmetry, $AI=DJ,$ $BI=CJ.$

four golden circles by Quang Tuan Bui, step 2

Triangles $CEJ$ and $MEI$ are similar so that $CE/EM=CJ/MI.$ From the foregoing discussion, $CE/EM=\sqrt{5}+1.$ Let $CJ=BI=x.$ Then $\displaystyle MI=\frac{1}{2}-x,$ giving an equation:


solving which we get $\displaystyle BI=x=\frac{3-\sqrt{5}}{2}.$ Now, $\displaystyle AI=1-x=\frac{\sqrt{5}-1}{2}$ and, as can be easily verified, $\displaystyle\frac{AI}{BI}=\phi.$ This proves #4.

Further, $\angle BEA=60^{\circ}$ so $\angle EGC=\angle EGF=120^{\circ}.$

four golden circles by Quang Tuan Bui, step 3

$\Delta EGC$ is isosceles, with $\angle CEG=30^{\circ},$ implying

$\begin{align}\displaystyle EG &=\frac{\frac{1}{2}CE}{\cos 30^{\circ}}\\ &=\frac{1}{2}\cdot \frac{\sqrt{3}}{2}(\sqrt{5}+1)\cdot\frac{2}{\sqrt{3}}\\ &=\frac{\sqrt{5}+1}{2}=\phi. \end{align}$

$FG=GC+CF=\phi +1=\phi^{2}$ such that $FG/EG=FG/CG=\phi,$ proving #1. We now possess all segment lengths to verify #3 directly.

The validity of #4 is the subject of the following


In a parallelogram $EHFG$ with angles of $60^{\circ}$ and $120^{\circ},$ if the sides relate in the Golden Ratio, so do the diagonals.


Using coordinates makes a 1-step proof.

four golden circles by Quang Tuan Bui, step 4

Let $E$ be the origin, $G=(1,0),$ $EH=\phi,$ and $\angle GEH=60^{\circ}.$ Then $\displaystyle H=\bigg(\frac{\phi}{2},\frac{\phi\sqrt{3}}{2}\bigg),$ implying

$\begin{align}\displaystyle GH^{2}&=\bigg(1-\frac{\phi}{2}\bigg)^{2}+\bigg(\frac{\phi\sqrt{3}}{2}\bigg)^{2}\\ &=1-\phi+\frac{\phi^{2}}{4}+\frac{3\phi^{2}}{4}\\ &=1-\phi+\phi^{2}\\ &=(1-\phi)+(1+\phi)=2. \end{align}$

Similarly, $EF^{2}=(1+\phi)+(1+\phi)=2(1+\phi)=2\phi^{2},$ so that $EF/GH=\phi.$

Note: In addition to #5 and #6, the diagram has several attractive properties (similar triangles, point collinearity, line concurrency, one angle being twice another, the angles between the diagonals being $60^{\circ}/120^{\circ},$ etc.) discovering which, along with proving #5 and $6 is left to the reader.

Golden Ratio

  1. Golden Ratio in Geometry
  2. Golden Ratio in Regular Pentagon
  3. Golden Ratio in an Irregular Pentagon
  4. Golden Ratio in a Irregular Pentagon II
  5. Inflection Points of Fourth Degree Polynomials
  6. Wythoff's Nim
  7. Inscribing a regular pentagon in a circle - and proving it
  8. Cosine of 36 degrees
  9. Continued Fractions
  10. Golden Window
  11. Golden Ratio and the Egyptian Triangle
  12. Golden Ratio by Compass Only
  13. Golden Ratio with a Rusty Compass
  14. From Equilateral Triangle and Square to Golden Ratio
  15. Golden Ratio and Midpoints
  16. Golden Section in Two Equilateral Triangles
  17. Golden Section in Two Equilateral Triangles, II
  18. Golden Ratio is Irrational
  19. Triangles with Sides in Geometric Progression
  20. Golden Ratio in Hexagon
  21. Golden Ratio in Equilateral Triangles
  22. Golden Ratio in Square
  23. Golden Ratio via van Obel's Theorem
  24. Golden Ratio in Circle - in Droves
  25. From 3 to Golden Ratio in Semicircle
  26. Another Golden Ratio in Semicircle
  27. Golden Ratio in Two Squares
  28. Golden Ratio in Two Equilateral Triangles
  29. Golden Ratio As a Mathematical Morsel
  30. Golden Ratio in Inscribed Equilateral Triangles
  31. Golden Ratio in a Rhombus
  32. Golden Ratio in Five Steps
  33. Between a Cross and a Square
  34. Four Golden Circles
  35. Golden Ratio in Mixtilinear Circles
  36. Golden Ratio With Two Equal Circles And a Line
  37. Golden Ratio in a Chain of Polygons, So to Speak
  38. Golden Ratio With Two Unequal Circles And a Line
  39. Golden Ratio In a 3x3 Square
  40. Golden Ratio In a 3x3 Square II
  41. Golden Ratio In Three Tangent Circles
  42. Golden Ratio In Right Isosceles Triangle
  43. Golden Ratio Poster
  44. Golden Ratio Next to the Poster
  45. Golden Ratio In Rectangles
  46. Golden Ratio In a 2x2 Square: Without And Within
  47. Golden Ratio With Two Unequal Circles And a Line II
  48. Golden Ratio in Equilateral and Right Isosceles Triangles
  49. Golden Ratio in a Butterfly Astride an Equilateral Triangle
  50. The Golden Pentacross
  51. 5-Step Construction of the Golden Ratio, One of Many
  52. Golden Ratio in 5-gon and 6-gon
  53. Golden Ratio in an Isosceles Trapezoid with a 60 degrees Angle
  54. Golden Ratio in Pentagon And Two Squares
  55. Golden Ratio in Pentagon And Three Triangles
  56. Golden Ratio in a Mutually Beneficial Relationship
  57. Star, Six Pentagons and Golden Ratio
  58. Rotating Square in Search of the Golden Ratio
  59. Cultivating Regular Pentagons
  60. Golden Ratio in an Isosceles Trapezoid with a 60 degrees Angle II
  61. More of Gloden Ratio in Equilateral Triangles
  62. Golden Ratio in Three Regular Pentagons
  63. Golden Ratio in Three Regular Pentagons II
  64. Golden Ratio in Wu Xing
  65. Golden Ratio In Three Circles And Common Secant

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