Circles in a Circular Segment

Many a sangaku come in clearly interrelated sequences, although their creation might have been spread of a few decades. We'll start with an example from [Fukagawa and Pedoe, Example 1.6].

Solution

References

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

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The tablet has been written in 1843, in the Aichi prefecture, and has since disappeared.

Solution

We have two show that the common tangent of the two circles O₂(r₂) and O₁(r₁) passes through the midpoint of the arc AB and also find the length of that tangent inside the circle O(r). The first task has been accomplished elsewhere by noting that the tangent serves as the radical axis of the two circles. Below we prove this result by computing the lengths of several line segments and using the converse of the Pythagorean theorem to show that ΔQTO₂ is right. By the same means we find the length of PQ.

With the reference to the diagram above, there are several right triangles to which we may fruitfully apply the Pythagorean theorem:

from the first we obtain (HO₁)² = 4r₁r₂ as in a related sangaku.

Substituting this into the second equation we get

r = r₂² / (r₂ - r₁).

Further, the similarity triangles HO₁O₂ and GTO₂

implies the proportion GT / HO₁ = TO₂ / O₁O₂, so that

GT = 2r₂√(r₁r₂) / (r₁ + r₂).

From the same triangles, GO₂ / HO₂ = O₂T / O₁O₂ giving

GO₂ = r₂(r₂ - r₁) / (r₁ + r₂).

Therefore,

GQ = 2r - r₂ - r₂(r₂ - r₁) / (r₁ + r₂).

One more application of the Pythagorean theorem helps determine QT from

QT² = GQ² + GT²

As the last step, we would like to show that ΔQO₂T is right (which would imply that QT is indeed tangent to the circles O₁(r₁) and O₂(r₂)). To this end, we are in a position to apply the converse of the Pythagorean theorem: suffice it to show that

QT² = O₂Q² - O₂T² = (r + r₂)² - r₂²,

which is left as an exercise.

From the orthogonality of QT and O₂T we see that triangles QO₂T and QSP are similar (∠QPS = 90° as subtended by the diameter QS.) Thus we get an additional proportion:

PQ / QT = QS / O₂T,

from which

PQ = QT · QS / O₂T = 4r₂²√(r₁r₂) / (r₂² - r₁²).

The next two problems are correspondingly [Fukagawa and Pedoe, #1.6.1] and [Fukagawa and Pedoe, #1.6.2]. Both tablets are lost. The first was written in 1857, Ibaragi prefecture, the second in 1806, Nagano prefecture.

The circles are tangent as shown. AB is the diameter of O(r). Find r₁, r₂ and r' in terms of r.

Solution: this is a combination of the above and another sangaku.

r₂ = r/2, r₁ = r/4, r' = r / (2 + √2)².

AB is no longer a diameter; the problem is to express r₁, r₂ and r' in terms of r and AB.

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
   • A Simple Solution to a Difficult Sangaku Problem
   • A Trigonometric 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. 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
    • Equal Incircle Theorem, Angela Drei's Proof
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
    • An Extension of a Sangaku with Touching Circles
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 © 1996-2018 Alexander Bogomolny