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, 1989
Write to:
Charles Babbage Research Center
P.O. Box 272, St. Norbert Postal Station
Winnipeg, MB
Canada R3V 1L6
Sangaku

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Copyright © 19962018 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:
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:
(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
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
Divide now by √R√R_{1}√R_{2} to obtain
Which is the desired formula.
Contact Front page Contents Geometry Up
Copyright © 19962018 Alexander Bogomolny