# Three Tangent Circles Sangaku

This is one of the rather more popular on the web Sangaku problems:

(See [Fukagawa & Pedoe, 1.1.2] where it is mentioned that the tablet survives in the Miyagai prefecture and is dated from 1892. There is a likely a misprint there as the authors refer for a solution to a publication from 1810.)

As some other Sangaku, this problem, too, requires nothing more than a few applications of the Pythagorean theorem. The main reason for its inclusion at the site is personal. I have only recently learned how to create in HTML the bar part of the symbol of square root. Because of the novelty, I am still enjoying doing that. As will be seen shortly, the Sangaku at hand provides an abundant opportunity to exercise the newly acquired skill.

### 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

## 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

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Copyright © 1996-2017 Alexander Bogomolny

As the diagram below shows we have three right triangles with the hypotenuses joining the centers of the three circles.

Using x and y to denote the horizontal distances between pairs of the circles, and R, R_{1}, R_{2} at their radii, the triangles have the following sides:

_{1}- R, x, R

_{1}+ R,

R

_{2}- R, y, R

_{2}+ R, and

R

_{2}- R

_{1}, x + y, R

_{2}+ R

_{1}.

The Pythagorean theorem then yields three equations in three unknowns, x, y, and R:

_{1}- R)

^{2}+ x

^{2}= (R

_{1}+ R)

^{2},

(R

_{2}- R)

^{2}+ y

^{2}= (R

_{2}+ R)

^{2}, and

(R

_{2}- R

_{1})

^{2}+ (x + y)

^{2}= (R

_{2}+ R

_{1})

^{2}.

After simplification the equations become

^{2}= 4R

_{1}R,

y

^{2}= 4R

_{2}R,

(x + y)

^{2}= 4R

_{1}R

_{2}.

Taking the square roots and substituting the first two into the third we get

_{1}+ √RR

_{2}= √R

_{1}√R

_{2}.

Divide now by √R√R_{1}√R_{2} to obtain

_{1}+ 1/√R

_{2}.

Which is the desired formula.

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Copyright © 1996-2017 Alexander Bogomolny

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