Three sangaku that require to determine the relative radii of the circles shown can be solved by a direct application of the Pythagorean theorem, see below, but if combined into a single (and a little amplified, sangaku-ready) configuration

reveal existence of possible relationships not obvious when they are studied separately. The configuration in fact, as was noted by L. Bankoff and C. W. Trigg quarter of a century before sangaku grew in popularity after a 1998 Scientific American article, is rich with surprises, the main being the numerous sightings of the famous 3:4:5 triangle.

The latter is most often referred to as the Rope-stretchers triangle and sometimes as the Egyptian triangle. (Both because of the belief that this simplest of the Pythagorean triangles was used by the ancient rope-stretchers in construction and, in particular, in construction of the Great Pyramids. Sometimes, however, the term Egyptian triangle is preserved for the one related to the dimensions of the pyramid of Cheops and the golden ratio.)

For convenience, assume the side of the square is 24 so that DF = FC = 12. Also, for simplicity, a circle with center X say, will be denoted (X). AH is tangent to (F) in G. Thus also FG = 12. The radius FG extends to meet BC in J and is perpendicular to AH.

HC = HG as tangents to (F) from H. For the same reason AG = AD = 24. FH bisects CFG and FA bisects its supplementary angle DFG.

AFH = 90^o

Then HB = 18, AH = 30, AK = 15, KG = 9, EK = 9, FK = 15, QK = 3, CJ = 16, FI = 8, EI = 16, AI = 20 (by the Pythagorean theorem), HJ = 10 (ΔFGK is similar to ΔJGH), BJ = 8, GJ = 8, FJ = 20, IQ = 4.

We see that triangles FCJ, HGJ, FGK, AKE, AEI, and ABH are all 3:4:5 triangles.

Let T be the intersection of AI extended with (A). Then AT = 24 and, since AI = 20, IT = 4 = IQ. Which says that the circle (I) of radius 4 is tangent to (A), (B), (F), and (E). If we extend AT further and let R on AT be on the vertical tangent SR to (I), then, for one, ΔIRS is similar to AEI and, hence, is also 3:4:5. Since IS = IQ = 4, IR = 20/3, SR = 16/3, RT = 8/3. If U is the intersection of SR and CD, then

In addition, if V is on BC and VC = RU, then VH = 10/3, RV = 8 and, by the Pythagorean theorem, HR = 26/3, so that again RX = 8/3. It follows that the circle (R) with radius 8/3 is tangent to (A), (I), (H) and CD. As an extra, it is also tangent to (W) with radius 3/2. ΔHVR is 5:12:13. The right triangle with vertical and horizontal legs and hypotenuse RW is 7:24:25.

References

1. L. Bankoff, C. W. Trigg, The Ubiquitous 3:4:5 Triangle, Math Magazine, v 47, n 2 (Mar., 1974), pp. 61-70

2. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

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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 49^th Degree Challenge
7. A Geometric Mean Sangaku
8. A Hard but Important Sangaku
9. A Restored Sangaku Problem
   • A better solution to a difficult 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. An Old Japanese Theorem
16. Archimedes Twins in the Edo Period
17. Arithmetic Mean Sangaku
18. Bottema Shatters Japan's Seclusion
19. Circles and Semicircles in Rectangle
20. Circles in a Circular Segment
21. Circles Lined on the Legs of a Right Triangle
22. Equal Incircles Theorem
23. Equilateral Triangle, Straight Line and Tangent Circles
24. Equilateral Triangles and Incircles in a Square
25. Five Incircles in a Square
26. Four Hinged Squares
27. Four Incircles in Equilateral Triangle
28. Gion Shrine Problem
29. Harmonic Mean Sangaku
30. Heron's Problem
31. In the Wasan Spirit
32. Incenters in Cyclic Quadrilateral
33. Japanese Art and Mathematics
34. Malfatti's Problem
35. Maximal Properties of the Pythagorean Relation
36. Neuberg Sangaku
37. Out of Pentagon Sangaku
38. Peacock Tail Sangaku
39. Pentagon Proportions Sangaku
40. Pythagoras and Vecten Break Japan's Isolation
41. Radius of a Circle by Paper Folding
42. Review of Sacred Mathematics
43. Sangaku à la V. Thebault
44. Sangaku and The Egyptian Triangle
45. Sangaku in a Square
46. Sangaku Iterations, Is it Wasan?
47. Sangaku with 8 Circles
48. Sangaku with Three Mixtilinear Circles
49. Sangaku with Versines
50. Sangakus with a Mixtilinear Circle
51. Sequences of Touching Circles
52. Square and Circle in a Gothic Cupola
53. Tangent Circles and an Isosceles Triangle
54. The Squinting Eyes Theorem
55. Steiner's Sangaku
56. Three Incircles In a Right Triangle
57. Three Squares and Two Ellipses
58. Three Tangent Circles Sangaku
59. Triangles, Squares and Areas from Temple Geometry
60. Two Arbelos, Two Chains
61. Two Circles in an Angle
62. Two Sangaku with Equal Incircles

Copyright © 1996-2009 Alexander Bogomolny

Let's, for example find the radius of (K) tangent to (A), (B) and AB. Actually we already saw that KE = KG = 9. If we wish to employ the Pythagorean theorem in ΔAEK, then denoting the radius of (K) as r we'll have

(24 - r)2 - r2 = 122,

which gives

576 - 48r = 144, or 48r = 432, r = 9.

Copyright © 1996-2009 Alexander Bogomolny