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, 1989
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Activities Contact Front page Contents Geometry
Copyright © 19962018 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
Sangaku

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Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny