Circles in a Circular Segment
Many a sangaku come in clearly interrelated sequences, although their creation might have been spread of a few decades. We'll start with an example from [Fukagawa and Pedoe, Example 1.6].
A chord AB divides a circle O(r) into two segments. A chain of three contact circles O_{3}(r_{1}), O_{2}(r_{2}), O_{1}(r_{1}) all touch AB and also O(r) internally, as shown in the figure. The tangent at the point of contact of the circle O_{2}(r_{2}) and O_{1}(r_{1}) meets the circle O(r) in P and Q. Show that Q is the midpoint of the arc AB, and find the length of PQ in terms r_{1} and r_{2}.
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989
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A chord AB divides a circle O(r) into two segments. A chain of three contact circles O_{3}(r_{1}), O_{2}(r_{2}), O_{1}(r_{1}) all touch AB and also O(r) internally, as shown in the figure. The tangent at the point of contact of the circle O_{2}(r_{2}) and O_{1}(r_{1}) meets the circle O(r) in P and Q. Show that Q is the midpoint of the arc AB, and find the length of PQ in terms r_{1} and r_{2}.
The tablet has been written in 1843, in the Aichi prefecture, and has since disappeared.
Solution
We have two show that the common tangent of the two circles O_{2}(r_{2}) and O_{1}(r_{1}) passes through the midpoint of the arc AB and also find the length of that tangent inside the circle O(r). The first task has been accomplished elsewhere by noting that the tangent serves as the radical axis of the two circles. Below we prove this result by computing the lengths of several line segments and using the converse of the Pythagorean theorem to show that ΔQTO_{2} is right. By the same means we find the length of PQ.
With the reference to the diagram above, there are several right triangles to which we may fruitfully apply the Pythagorean theorem:
ΔO_{2}HO_{1}:  (r_{2}  r_{1})² + (HO_{1})² = (r_{1} + r_{2})²,  
ΔHOO_{1}:  (r  2r_{2} + r_{1})² + (HO_{1})² = (r  r_{1})², 
from the first we obtain (HO_{1})² = 4r_{1}r_{2} as in a related sangaku.
Substituting this into the second equation we get
r = r_{2}² / (r_{2}  r_{1}).
Further, the similarity triangles HO_{1}O_{2} and GTO_{2}
implies the proportionGT = 2r_{2}√(r_{1}r_{2}) / (r_{1} + r_{2}).
From the same triangles, GO_{2} / HO_{2} = O_{2}T / O_{1}O_{2} giving
GO_{2} = r_{2}(r_{2}  r_{1}) / (r_{1} + r_{2}).
Therefore,
GQ = 2r  r_{2}  r_{2}(r_{2}  r_{1}) / (r_{1} + r_{2}).
One more application of the Pythagorean theorem helps determine QT from
QT² = GQ² + GT²
As the last step, we would like to show that ΔQO_{2}T is right (which would imply that QT is indeed tangent to the circles O_{1}(r_{1}) and O_{2}(r_{2})). To this end, we are in a position to apply the converse of the Pythagorean theorem: suffice it to show that
QT² = O_{2}Q²  O_{2}T² = (r + r_{2})²  r_{2}²,
which is left as an exercise.
From the orthogonality of QT and O_{2}T we see that triangles QO_{2}T and QSP are similar
PQ / QT = QS / O_{2}T,
from which
PQ = QT · QS / O_{2}T = 4r_{2}²√(r_{1}r_{2}) / (r_{2}²  r_{1}²).
The next two problems are correspondingly [Fukagawa and Pedoe, #1.6.1] and [Fukagawa and Pedoe, #1.6.2]. Both tablets are lost. The first was written in 1857, Ibaragi prefecture, the second in 1806, Nagano prefecture.
The circles are tangent as shown. AB is the diameter of O(r). Find r_{1}, r_{2} and r' in terms of r.
Solution: this is a combination of the above and another sangaku.
r_{2} = r/2, r_{1} = r/4, r' = r / (2 + √2)².
AB is no longer a diameter; the problem is to express r_{1}, r_{2} and r' in terms of r and AB.
Sangaku

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Copyright © 19962018 Alexander Bogomolny