3-4-5 Triangle by a Kid

It is often said that the practice of hanging sangaku was so common that even women and children took part in that activity. While there is a reasonable doubt as to the popularity of sangaku, there is no question that some were hung by very young children. The one below was proposed by Sato Naosue, age thirteen, and hung in 1847 in the Akahagi Kannon temple in Ichinoseki city.

Two pink circles of radius $r$ and two white circles of radius $t$ are inscribed in a square, as shown. The square itself is inscribed in a large triangle and, as illustrated, two circles of radii $r$ and $R$ are inscribed in the small triangles outside the square. Show that $R = 2t$.

Solution

References

  1. H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008, pp. 104-105

Sangaku

  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
  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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Copyright © 1996-2017 Alexander Bogomolny

Obviously, the side of the square is of length $4r.$ One application of the Pythagorean theorem (in the dashed triangle) shows that $r = 3t/2.$ Use the Pythagorean theorem again in the small upper triangle:

Let $a$ be the small length of that triangle and $c$ its hypotenuse. On the other hand, the area $S$ of the small triangle can be computed in two ways:

$2S=r(a+c+4r)=4r\cdot a,$

such that $a+c+4r=4a;$ from which $c=3a-4r.$ Substitute that into the Pythagorean identity - $a^2+(4r)^2=c^2$ - to get the equation

$a^2+16r^2=(3a-4r)^2,$

solving which gives $a=3r.$

It may come as a surprise that the triangle turns out to be the famous 3-4-5 or Egyptian triangle with the short side equal to $3r.$ From the similarity of the three triangles, all of them have the proportions 3-4-5 which leads to $4r = 3R.$ And finally, $R = 4r/3 = 2t.$

|Contact| |Front page| |Contents| |Geometry| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

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