# 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 O_{1}, O_{2}, and O_{3} are collinear points and the circles O_{1}(r), O_{2}(r), and O_{3}(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 O_{1}(r) and O_{3}(r). Show that

What if applet does not run? |

### References

- H. Fukagawa, D. Pedoe,
*Japanese Temple Geometry Problems*, The Charles Babbage Research Center, Winnipeg, 1989Write 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 O_{1}, O_{2}, and O_{3} are collinear points and the circles O_{1}(r), O_{2}(r), and O_{3}(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 O_{1}(r) and O_{3}(r). Show that

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

So N circles of radius r are located on a chord, with midpoint O'; circle O(R) touches circles O_{1}(r) and O_{N}(r) internally; chord PQ is internally tangent to O_{1}(r) and O_{N}(r). Let W be the point of tangency of O(R) and O_{N}(r), S be the point of tangency of PQ and O_{N}(r). Let d denote the distance between O and O' and α be the angle QO'O_{N}.

In ΔOO'O_{N}, (OO')² + (O'O_{N})² = (OO_{N})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'O_{N}S, sineα = 1/(N - 1) such that

(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,

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 = 2√R² - [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 _{N} = ρ - a/N_{N}ρ = ρ + b/N_{N})_{N}ρ)

## Sangaku

- Sangaku: Reflections on the Phenomenon
- Critique of My View and a Response
- 1 + 27 = 12 + 16 Sangaku
- 3-4-5 Triangle by a Kid
- 7 = 2 + 5 Sangaku
- A 49
^{th}Degree Challenge - A Geometric Mean Sangaku
- A Hard but Important Sangaku
- A Restored Sangaku Problem
- A Sangaku: Two Unrelated Circles
- A Sangaku by a Teen
- A Sangaku Follow-Up on an Archimedes' Lemma
- A Sangaku with an Egyptian Attachment
- A Sangaku with Many Circles and Some
- A Sushi Morsel
- An Old Japanese Theorem
- Archimedes Twins in the Edo Period
- Arithmetic Mean Sangaku
- Bottema Shatters Japan's Seclusion
- Chain of Circles on a Chord
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
- Circles Lined on the Legs of a Right Triangle
- Equal Incircles Theorem
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
- Five Incircles in a Square
- Four Hinged Squares
- Four Incircles in Equilateral Triangle
- Gion Shrine Problem
- Harmonic Mean Sangaku
- Heron's Problem
- In the Wasan Spirit
- Incenters in Cyclic Quadrilateral
- Japanese Art and Mathematics
- Malfatti's Problem
- Maximal Properties of the Pythagorean Relation
- Neuberg Sangaku
- Out of Pentagon Sangaku
- Peacock Tail Sangaku
- Pentagon Proportions Sangaku
- Proportions in Square
- Pythagoras and Vecten Break Japan's Isolation
- Radius of a Circle by Paper Folding
- Review of Sacred Mathematics
- Sangaku à la V. Thebault
- Sangaku and The Egyptian Triangle
- Sangaku in a Square
- Sangaku Iterations, Is it Wasan?
- Sangaku with 8 Circles
- Sangaku with Angle between a Tangent and a Chord
- Sangaku with Quadratic Optimization
- Sangaku with Three Mixtilinear Circles
- Sangaku with Versines
- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
- Square and Circle in a Gothic Cupola
- Steiner's Sangaku
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Three Incircles In a Right Triangle
- Three Squares and Two Ellipses
- Three Tangent Circles Sangaku
- Triangles, Squares and Areas from Temple Geometry
- Two Arbelos, Two Chains
- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku

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

Copyright © 1996-2018 Alexander Bogomolny

63241499 |