# Three Tangent Circles Sangaku

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

Given three circles tangent to each other and to a straight line, express the radius of the middle circle via the radii of the other two.

(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.

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

## Sangaku

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, R1, R2 at their radii, the triangles have the following sides:

R1 - R, x, R1 + R,
R2 - R, y, R2 + R, and
R2 - R1, x + y, R2 + R1.

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

(R1 - R)2 + x2 = (R1 + R)2,
(R2 - R)2 + y2 = (R2 + R)2, and
(R2 - R1)2 + (x + y)2 = (R2 + R1)2.

After simplification the equations become

x2 = 4R1R,
y2 = 4R2R,
(x + y)2 = 4R1R2.

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

RR1 + RR2 = R1R2.

Divide now by RR1R2 to obtain

1/R = 1/R1 + 1/R2.

Which is the desired formula.