Chain of Circles on a Chord

Here's Problem 1.3.3 from [Fukagawa & Pedoe]. The tablet, written in 1871, survives in the Ibaragi prefecture.

Points O1, O2, and O3 are collinear points and the circles O1(r), O2(r), and O3(r) touch each other, the first touching the second and the second the third. The circle O(R) circumscribes the three given circles, touching the first and the third internally. The chord PQ of this circle is an internal common tangent to the circles O1(r) and O3(r). Show that PQ = R + 3r.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Points O1, O2, and O3 are collinear points and the circles O1(r), O2(r), and O3(r) touch each other, the first touching the second and the second the third. The circle O(R) circumscribes the three given circles, touching the first and the third internally. The chord PQ of this circle is an internal common tangent to the circles O1(r) and O3(r). Show that PQ = R + 3r.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

We shall generalize the problem a little. Instead of 3 circles we shall consider a chain of N touching circles, with N ≥ 3. This does not make the problem much more difficult.

So N circles of radius r are located on a chord, with midpoint O'; circle O(R) touches circles O1(r) and ON(r) internally; chord PQ is internally tangent to O1(r) and ON(r). Let W be the point of tangency of O(R) and ON(r), S be the point of tangency of PQ and ON(r). Let d denote the distance between O and O' and α be the angle QO'ON.

In ΔOO'ON, (OO')² + (O'ON)² = (OON)², which, in our notations, means

d² + [(N - 1)r]² = (R - r)².

As an equation for r, this can be rewritten as

((N - 1)² - 1)r² + 2Rr - (R² - d²) = 0.

Which leads to

r = [-R ± [R² + ((N - 1)² - 1)(R² - d²)] / ((N - 1)² - 1).

Choosing "+' in front of the radical, we obtain

N(N - 2)r = -R + [(N - 1)²R² - ((N - 1)² - 1)d².

From the right triangle O'ONS, sineα = 1/(N - 1) such that cos²α = ((N - 1)² - 1) / (N - 1)². So we may shorten (1):

(1') N(N - 2)r + R = (N - 1)[R² - [d·cosα]².

In ΔOO'Q, by the Law of Cosines,

OQ² = d² - 2d·O'Q·cos(90° + α) + (O'Q)².

With O'Q = s, we get a quadratic equation

s² + 2d·s·sin(α) + (d² - R²).

Similarly, with O'P = t, we have in ΔOO'P, ∠OO'P = 90° - α and

t² - 2d·s·sin(α) + (d² - R²).

Solving for s and t gives

 s= -d·sin(α) ± [d·sin(α)]² + (R² - d²).
 t= +d·sin(α) ± [d·sin(α)]² + (R² - d²).

Naturally, R > d which forces the sign "+" before the radicals. We can now find PQ = s + t:

(2) PQ = 2R² - [d·cos(α)]²

Luckily, this is the same radical as in (1'). Eliminating the radical gives

PQ = 2 (R + N(N - 2)r) / (N - 1),

which, for N = 3, gives the required result.

Now, let's denote ρ = Nr, which is the radius of the circle (the dashed circle that appears when the box "extra" is checked) that has the small circles lined up on its diameter. Then we have

(3) PQ = 2 (R + (N - 2)ρ) / (N - 1),

or, for N = 3, PQ = R + ρ. It's a pity we need a more complicated formula (3) in the general case.

A very enticing observation has been made in an article by J. Marshall Unger (Problem #5). Imagine the dynamics of moving from circle O'(ρ) to the circle O(R) while the line PQ and the small circles remain untouched. P'Q' is the corresponding chord in O'(ρ). Let a and b be the distances between the top and, respectively, bottom points of circles O'(ρ) and O(R) so that the change of radius is (b - a)/2: R - ρ = (a - b)/2. Then the change in the chord length is PQ - P'Q' = (b - a)/N. (I assume b > a). Moreover, if we introduce ρN = ρ - a/N and Nρ = ρ + b/N then, as illustrated in the applet, the two circles O'(ρN) and O'(Nρ) - the dotted circles in the applet - meet O(R) at exactly the points Q and P!

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

65814156