Four Incircles in an Equilateral Triangle, Third Solution

Elsewhere we described and solved the following sangaku:

In an equilateral triangle \(ABC\) the lines \(BEF\), \(CFD\), \(ADE\) are drawn making equal angles with \(AB\), \(BC\), \(CA\), respectively, forming the triangle \(DEF\), and so that the radius of the incircle of triangle \(DEF\) is equal to the radii of the incircles of triangles $ABE,$ $BCF,$ and $ACD.$

four incircles sangaku

Find \(DE\) in terms of \(AB\).

The solution below is by J. John Samuel (7 MAy, 2014). Like George Zettler, John first establishes the basic facts, viz.,

  1. The area $A$ of a triangle with the semiperimeter $p$ and the inradius $r$ equals $A=pr.$
  2. In an equilateral triangle the side length $a$ and the inradius $r$ are related through $a = 2\sqrt{3}r.$
  3. With the reference to the above diagram, in $\Delta ABE,$ the altitude $h$ from vertex $B$ equals $\displaystyle h=\frac{\sqrt{3}}{2}BE.$

Next, with a change of notations reflected in the diagram below,

four incircles sangaku - J. John Samuel's proof

John derives the expression

$\displaystyle \frac{S_{o}^{2}}{S_{i}^{2}}=3k(k+1)+1,$

where $S_{o}$ and $S_{i}$ are the side lengths of the $o$uter and the $i$nner equilateral triangles. The easiest way to see that is to observe that, say, in $\Delta ABE,$ the Law of Cosines gives

$S_{o}^{2}=((kS_{i})^{2}+(kS_{i})((k+1)S_{i})+((k+1)S_{i})^{2}=(S_{i})^{2}(k^{2}+k(k+1)+(k+1)^{2}).$

The final step is illustrated by a diagram and a moving observation:

four incircles sangaku - J. John Samuel's proof, the final step

We may find the relationship between $S_{o}$ and $S_{i}$ in yet another way, through the area formula $A=pr:$

$\displaystyle\begin{align} Area(\Delta ABE) &= \frac{1}{2}(S_{o}+kS_{i}+(k+1)S_{i})R\\ &=\frac{1}{2}\frac{S_{i}}{2\sqrt{3}}(S_{o}+(2k+1)S_{i})\\ &=\frac{S_{i}}{4\sqrt{3}}(S_{o}+(2k+1)S_{i}). \end{align}$

On the other hand,

$\displaystyle\begin{align} Area(\Delta ABE) &= \frac{1}{2}h\cdot AE\\ &=\frac{1}{2}\frac{\sqrt{3}}{2}(kS_{i})((k+1)S_{i})\\ &=\frac{\sqrt{3}}{4}S_{i}^{2}k(k+1). \end{align}$

Comparing the two expressions, we obtain $S_{o}=S_{i}(3k^{2}+k-1),$ which leads to an equation for k:

$\displaystyle\frac{S_{o}^{2}}{S_{i}^{2}}=(3k^{2}+k-1)^{2}=3k(k+1)+1.$

This is a fourth degree equation. Undeterred, we proceed to simplify:

$\displaystyle\begin{align} 3k(k+1)+1 &= (3k^{2}+k-1)^{2}\\ &=9k^{4}+6k^{3}-5k^{2}-2k+1, \end{align}$

or, $9k^{4}+6k^{3}-8k^{2}-5k=0.$ $k=0,$ not fitting the context of the problem can be cancelled out: $9k^{3}+6k^{2}-8k-5=0.$ In addition, we may notice (and John does) that $-9+6+8-5=0,$ implying that $k=-1$ is another irrelevant root of the equation. Dividing by $k+1$ leaves $9k^{2}-3k-5=0,$ with the roots $\displaystyle\frac{3\pm\sqrt{189}}{18}=\frac{1\pm\sqrt{21}}{6}.$ Dropping the negative root, we are left with the only suitable solution

$\displaystyle k=\frac{1+\sqrt{21}}{6}. $

Now we are in a position to solve the problem:

$\displaystyle \frac{S_{o}}{S_{i}}=3k^{2}+k-1=\frac{3+\sqrt{21}}{3}. $

From this

$\displaystyle \frac{S_{i}}{S_{o}}=\frac{3}{3+\sqrt{21}}=\frac{3(\sqrt{21}-3)}{21-9}=\frac{\sqrt{21}-3}{4}. $

This agrees with the solutions of George Zettler and Angela Drei.

  1. Sangaku: Reflections on the Phenomenon
  2. Critique of My View and a Response
  3. 1 + 27 = 12 + 16 Sangaku
  4. 3-4-5 Triangle by a Kid
  5. 7 = 2 + 5 Sangaku
  6. A 49th Degree Challenge
  7. A Geometric Mean Sangaku
  8. A Hard but Important Sangaku
  9. A Restored Sangaku Problem
  10. A Sangaku: Two Unrelated Circles
  11. A Sangaku by a Teen
  12. A Sangaku Follow-Up on an Archimedes' Lemma
  13. A Sangaku with an Egyptian Attachment
  14. A Sangaku with Many Circles and Some
  15. A Sushi Morsel
  16. An Old Japanese Theorem
  17. Archimedes Twins in the Edo Period
  18. Arithmetic Mean Sangaku
  19. Bottema Shatters Japan's Seclusion
  20. Chain of Circles on a Chord
  21. Circles and Semicircles in Rectangle
  22. Circles in a Circular Segment
  23. Circles Lined on the Legs of a Right Triangle
  24. Equal Incircles Theorem
  25. Equilateral Triangle, Straight Line and Tangent Circles
  26. Equilateral Triangles and Incircles in a Square
  27. Five Incircles in a Square
  28. Four Hinged Squares
  29. Four Incircles in Equilateral Triangle
    • Four Incircles in an Equilateral Triangle, a Sangaku
  30. Gion Shrine Problem
  31. Harmonic Mean Sangaku
  32. Heron's Problem
  33. In the Wasan Spirit
  34. Incenters in Cyclic Quadrilateral
  35. Japanese Art and Mathematics
  36. Malfatti's Problem
  37. Maximal Properties of the Pythagorean Relation
  38. Neuberg Sangaku
  39. Out of Pentagon Sangaku
  40. Peacock Tail Sangaku
  41. Pentagon Proportions Sangaku
  42. Proportions in Square
  43. Pythagoras and Vecten Break Japan's Isolation
  44. Radius of a Circle by Paper Folding
  45. Review of Sacred Mathematics
  46. Sangaku à la V. Thebault
  47. Sangaku and The Egyptian Triangle
  48. Sangaku in a Square
  49. Sangaku Iterations, Is it Wasan?
  50. Sangaku with 8 Circles
  51. Sangaku with Angle between a Tangent and a Chord
  52. Sangaku with Quadratic Optimization
  53. Sangaku with Three Mixtilinear Circles
  54. Sangaku with Versines
  55. Sangakus with a Mixtilinear Circle
  56. Sequences of Touching Circles
  57. Square and Circle in a Gothic Cupola
  58. Steiner's Sangaku
  59. Tangent Circles and an Isosceles Triangle
  60. The Squinting Eyes Theorem
  61. Three Incircles In a Right Triangle
  62. Three Squares and Two Ellipses
  63. Three Tangent Circles Sangaku
  64. Triangles, Squares and Areas from Temple Geometry
  65. Two Arbelos, Two Chains
  66. Two Circles in an Angle
  67. Two Sangaku with Equal Incircles
  68. Another Sangaku in Square
  69. Sangaku via Peru
  70. FJG Capitan's Sangaku

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