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.
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.
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 = √R1√R2.
Divide now by √R√R1√R2 to obtain
1/√R = 1/√R1 + 1/√R2.
Which is the desired formula.
|Contact|
|Front page|
|Contents|
|Geometry|
|Up|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
