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].

A chord AB divides a circle O(r) into two segments. A chain of three contact circles O3(r1), O2(r2), O1(r1) all touch AB and also O(r) internally, as shown in the figure. The tangent at the point of contact of the circle O2(r2) and O1(r1) meets the circle O(r) in P and Q. Show that Q is the midpoint of the arc AB, and find the length of PQ in terms r1 and r2.

Solution

References

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

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6

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A chord AB divides a circle O(r) into two segments. A chain of three contact circles O3(r1), O2(r2), O1(r1) all touch AB and also O(r) internally, as shown in the figure. The tangent at the point of contact of the circle O2(r2) and O1(r1) meets the circle O(r) in P and Q. Show that Q is the midpoint of the arc AB, and find the length of PQ in terms r1 and r2.

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:

ΔO2HO1:  (r2 - r1)² + (HO1)² = (r1 + r2)²,
ΔHOO1:  (r - 2r2 + r1)² + (HO1)² = (r - r1)²,

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 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 © 1996-2018 Alexander Bogomolny

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