A Restored Sangaku Problem

The sangaku discussed below was originally hung [Fukagawa and Rothman, pp. 194-195] in 1806 in the Atsuta shrine by Ehara Masanori, a disciple of Kusaka Makoto (1764-1833). The tablet, which contained only this problem and no solution, was subsequently lost. However, at an unknown date the mathematician Kitagawa Moko (1763-1839) visited the shrine and recorded the sangaku in his note "Kyuka Sankei," or "Nine Flowers Mathematics," along with his solution. More recently the shrine constructed a replica from Kitagawa's manuscript. Fukagawa and Rothman reproduce Kitagawa's solution with a warning: This may be the most involved exploitation of the Pythagorean theorem you have ever seen.

Chapter 6, problem 3, Fukagawa & Rothman

The problem is this: ΔABC is isosceles. Two lines BD and CH divide it into three smaller triangles, each of which circumscribes a circle of the same radius r. Find r in terms of CH.

Following [Fukagawa and Rothman, pp. 212-216], I shall give Kitagawa's solution and then, on a separate page, a solution by J. M. Unger.

Chapter 6, problem 3, Fukagawa & Rothman. Inner isosceles triangle

First, notice from the figure above that triangle CDB is isosceles. Therefore let b = BC = CD, as indicated. We also let a = BD and k = CH. Next, as shown below, mark with a compass two points D' on AB and D'' on AC such that BD = BD' = CD'' = a. Because ΔABC is isosceles, D'T' = D''T'' and by construction D'T' = DT; consequently DT = D''T''.

Chapter 6, problem 3, Fukagawa & Rothman, just two circles

T and T'' are equidistant from D and therefore DT'' = D''T''. Defining s = DT = DT'' we can write

(1) s = (CD'' - CD) / 2 = (a - b) / 2.

Further, by examining the tangents in the figure above, we see that

(2) b = (k - r) + (a/2 - r) = a/2 + k - 2r.

Hence, from equation (1)

(3) s = a / 4 - (k - 2r) / 2.

Also, with the Pythagorean theorem, from either ΔBCH or ΔCDH, b² = (a/2)² + k². Squaring equation (2) and equating the two expressions quickly yields

(4) a = 4r(k - r) / (k - 2r).

Next, as shown below, inscribe a circle of radius R in triangle BCD.

Chapter 6, problem 3, Fukagawa & Rothman, with a bog incircle

With the help of the figure it is easy to show that

(5) CQ = (k - R)² - R² = k² - 2Rk = b - a/2 = k - 2r,

where the last equality follows from equation (2). Squaring equation (5) gives 2Rk = (4kr) - 4r², or

(6) R = 2r (k - r) / k.

The strategy is now to eliminate R in favor of r and k. Applying the Pythagorean theorem directly to the figure below gives

(R + r)² + (a/2 - s)² = (R - r)² + (s + a/2)²,

Chapter 6, problem 3, Fukagawa & Rothman, final Pythagoras

or, after simplifying,

(7) 2Rr = as.

The problem is almost solved. We now replace R on the left of equation (7) by equation (6) and as on the right by the expressions in equations (3) and (4) to get

4r²(k - r) / k = 4r(k - r)/(k - 2r) × [(r(k - r)/(k - 2r) - (k - 2r)/2],

which gives the cubic

k³ - 4rk² - 2kr² + 8r³ = (k² - 2r²)(k - 4r) = 0.

From any of the figures, the relevant root is k = 4r, and thus we have the final result

r = CH / 4.

References

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

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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