Two Arbelos, Two Chains
There is a Pappus' chain in arbelos, the shoemaker's knife, and, vacuously, two of them in two cobbler implements. In a particular sangaku documented as 1.8.5 in the collection by Fukagawa and Pedoe the two devices have much on common which induces a relationship between the circles in their chains:
Points T, A, B, C are collinear and AB = BC = CT = 2r. Circles S_{1}(3r) and S_{2}(2r) are drawn on AT and BT respectively as diameters. We consider the chain of contact circles O_{i}(r_{i}),
t_{n} / (t_{n} / r_{n}  3) = 2r / 13.
Note that the required identity is likely to be faulty, for it's quite obvious from the construction that
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Points T, A, B, C are collinear and AB = BC = CT = 2r. Circles S_{1}(3r) and S_{2}(2r) are drawn on AT and BT respectively as diameters. We consider the chain of contact circles O_{i}(r_{i}),
t_{n} / (t_{n} / r_{n}  3) = 2r / 13.
This sangaku is said to be written in 1842 in the Nagano prefecture. The table has since disappeared.
Solution
We make use of a useful formula twice getting, for the small arbelos
t_{n} = 2r / (n² + 2)
and, for the big one,
r_{n} = 3r·1/2 / ((n/2)² + 1/2 + 1) = 6r / (n² + 6).
In order to derive at anything resembling 2r/13, we have to somehow get rid of the dependency on n. The easiest way to do that is to pass to the reciprocals:
1 / t_{n} = (n² + 2) / 2r and
1 / r_{n} = (n² + 6) / 6r
Obviously,
3 / r_{n}  1 / t_{n} = 4 / 2r = 2 / r,
independent of n. The latter can be rewritten as
1 / r_{n} · (3  r_{n} / t_{n}) = 2 / r,
or
r_{n} / (3  r_{n} / t_{n}) = r / 2.
Perhaps, this is what the tablet was intended for.
Sangaku

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Copyright © 19962018 Alexander Bogomolny