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 O₁, O₂, and O₃ are collinear points and the circles O₁(r), O₂(r), and O₃(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 O₁(r) and O₃(r). Show that PQ = R + 3r.

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Solution

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

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We shall generalize the problem a little. Instead of 3 circles we shall consider a chain of N touching circles, with N ≥ 3. This does not make the problem much more difficult.

So N circles of radius r are located on a chord, with midpoint O'; circle O(R) touches circles O₁(r) and O_N(r) internally; chord PQ is internally tangent to O₁(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_NS, sineα = 1/(N - 1) such that cos²α = ((N - 1)² - 1) / (N - 1)². So we may shorten (1):

(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, ∠OO'P = 90° - α and

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 (b - a)/2: R - ρ = (a - b)/2. Then the change in the chord length is PQ - P'Q' = (b - a)/N. (I assume b > a). Moreover, if we introduce ρ_N = ρ - a/N and _Nρ = ρ + b/N then, as illustrated in the applet, the two circles O'(ρ_N) and O'(_Nρ) - the dotted circles in the applet - meet O(R) at exactly the points Q and P!

Sangaku

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