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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

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

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

Copyright © 1996-2008 Alexander Bogomolny

28698241Page copy protected against web site content infringement by Copyscape


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