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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 (1764-1839). 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. Circles and Semicircles in Rectangle
  21. Circles in a Circular Segment
  22. Circles Lined on the Legs of a Right Triangle
  23. Equal Incircles Theorem
  24. Equilateral Triangle, Straight Line and Tangent Circles
  25. Equilateral Triangles and Incircles in a Square
  26. Five Incircles in a Square
  27. Four Hinged Squares
  28. Four Incircles in Equilateral Triangle
  29. Gion Shrine Problem
  30. Harmonic Mean Sangaku
  31. Heron's Problem
  32. In the Wasan Spirit
  33. Incenters in Cyclic Quadrilateral
  34. Japanese Art and Mathematics
  35. Malfatti's Problem
  36. Maximal Properties of the Pythagorean Relation
  37. Neuberg Sangaku
  38. Out of Pentagon Sangaku
  39. Peacock Tail Sangaku
  40. Pentagon Proportions Sangaku
  41. Pythagoras and Vecten Break Japan's Isolation
  42. Radius of a Circle by Paper Folding
  43. Review of Sacred Mathematics
  44. Sangaku à la V. Thebault
  45. Sangaku and The Egyptian Triangle
  46. Sangaku in a Square
  47. Sangaku Iterations, Is it Wasan?
  48. Sangaku with 8 Circles
  49. Sangaku with Three Mixtilinear Circles
  50. Sangaku with Versines
  51. Sangakus with a Mixtilinear Circle
  52. Sequences of Touching Circles
  53. Square and Circle in a Gothic Cupola
  54. Tangent Circles and an Isosceles Triangle
  55. The Squinting Eyes Theorem
  56. Steiner's Sangaku
  57. Three Incircles In a Right Triangle
  58. Three Squares and Two Ellipses
  59. Three Tangent Circles Sangaku
  60. Triangles, Squares and Areas from Temple Geometry
  61. Two Arbelos, Two Chains
  62. Two Circles in an Angle
  63. Two Sangaku with Equal Incircles

Copyright © 1996-2009 Alexander Bogomolny

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