Cultivating Regular Pentagons


Cultivating Regular Pentagons, problem

Solution 1

First note that, being an isosceles trapezoid, $CDFH\,$ is cyclic. Let $I'\,$ be the point where circle $(CDFH)\,$ meets $FG.$

Cultivating Regular Pentagons, #2

In the regular pentagon $ABCDE,\,$ $AC\parallel DE,\,$ and let the extension of $AC\,$ cross $FG\,$ in $I''.\,$ We'll show that $I'=I''=I.$

Trapezoid $CDFI''\,$ is isosceles, hence, cyclic. $(CDFI'')=(CDF)=(CDFH)\,$ from which $I''=I'.\,$ Further, in the isosceles trapezoid $CDFI''\,$ $CF=DI'';\,$ in the isosceles trapezoid $CDFH\,$ $CF=DH.\,$ By transitivity, $DI''=CF=DH,\,$ such that $I''\in D(H)\,$ and, subsequently, $I''=I.$

To solve the original problem, suffice it to note that $AC\,$ divides $BE\,$ in the golden ratio. Thus, by Thales' theorem, it is also true for $FG:$

Cultivating Regular Pentagons, #3

Solution 2

Let $\beta=72^o\,$ and the side of the smaller pentagon be $p.\,$ Let $\displaystyle k=\frac{\varphi}{1+\varphi}.$

The side of the larger pentagon is then $q=2p\cos(\beta/2).\,$ With origin at $D\,$ and the $X\,$ axis parallel to $CD,\,$ the coordinates of the relevant points are:

$\displaystyle \begin{align} F&\rightarrow \left((p+q)\cos\beta,(p+q)\sin\beta\right), \\ H&\rightarrow \left(-p-(p+q)\cos\beta,(p+q)\sin\beta\right), \\ G&\rightarrow \left((p+q)\cos\beta+q\cos 2\beta,(p+q)\sin\beta+q\sin 2\beta\right), \\ I&\rightarrow \left((p+q)\cos\beta+kq\cos 2\beta,(p+q)\sin\beta+kq\sin 2\beta\right). \end{align}$

Both, $H\,$ and $I\,$ being on the circle centered at $D\,$ means

$\displaystyle \begin{align} &|I|^2=|H|^2 \\ &[(p+q)\cos\beta+kq\cos 2\beta]^2+[(p+q)\sin\beta+kq\sin 2\beta]^2 = \\ &\qquad\qquad\qquad[-p-(p+q)\cos\beta]^2+[(p+q)\sin\beta]^2 \\ &k^2q^2+2kq(p+q)\cos\beta-p^2-2p(p+q)\cos\beta=0 \\ &4k^2\cos^2\left(\frac{\beta}{2}\right)+4\cos\left(\frac{\beta}{2}\right)k\left[1+2\cos\left(\frac{\beta}{2}\right)\right]\cos\beta-1 \\ &\qquad\qquad\qquad-2\left[1+2\cos\left(\frac{\beta}{2}\right)\right]\cos\beta=0 \end{align}$

Plugging in the values:

$\displaystyle \begin{align} \cos\left(\frac{\beta}{2}\right)&= \cos~36^o=\frac{\sqrt{5}+1}{4} \\ \cos\beta&= \cos~72^o=\frac{\sqrt{5}-1}{4}, \end{align}$


$\displaystyle 2(\sqrt{5}+3)k^2+(\sqrt{5}+1)^2k-2(\sqrt{5}+3)=0$

with roots

$\displaystyle k=\left\{\left(\frac{\sqrt{5}-1}{2}\right),-\left(\frac{\sqrt{5}+1}{2}\right)\right\}.$

Choosing the physical root,

$\displaystyle \begin{align} \frac{\varphi}{1+\varphi}&= \frac{\sqrt{5}-1}{2}, \\ \varphi&= \frac{\sqrt{5}+1}{2}. \end{align}$


We may continue building a pyramid of regular pentagons, as shown below:

Cultivating Regular Pentagons, #4

As above, the circles $D(M),\,\,$ $D(P),\,\,$ $D(S),\,\,$ etc., meet $AC\,\,\,$ on the sides of the corresponding pentagons. If $N=AC\cap KL=D(M)\cap KL,\,\,$ then $\displaystyle \frac{LN}{KN}=\varphi,\,\,$ the Golden Ratio.


The problem has been kindly posted by Tran Quang Hung at the CutTheKnotMath facebook page. GeoGebra made it easy to draw and investigate the construction. Solution 2 is by Amit Itagi.


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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