A Sangaku with Many Circles and Some: What Is This About?
A Mathematical Droodle
What if applet does not run? |
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Copyright © 1996-2018 Alexander Bogomolny
The applet purports to suggest the following 1837 sangaku [Temple Geometry, 1.5.7] from the Aichi prefecture:
AB is a diameter of O(r) and forms two semicircles. In one of these the equal circles O_{1}(r_{1}) and O'_{1}(r_{1}) touch AB and also O(r) internally. The circles O_{2}(r_{1}) and O'_{2}(r_{1}) are mirror images of these two circles in AB. The circle O(r_{3}) touches the 4 circles we have just described externally, and the circle O_{4}(r_{4}) touches O_{1}(r_{1}) and O'_{1}(r_{1}) externally and O(r) internally. Show that
In the particular case where r_{3} = r/9, show that
where O_{5}(r_{5}) has interior contact, as shown, with both O_{1}(r_{1}) and O'_{1}(r_{1}). |
We'll proceed a step at a time. First, let X be the points of contact of O_{1}(r_{1}) with AB and x = OX.
By the Pythagorean theorem,
(r - r_{1})² = x² + (r_{1})². |
From which
(1) | r_{1} = (r² - x²) / 2r. |
Next we find r_{3}:
(r_{3} + r_{1})² = x² + (r_{1})². |
implying a quadratic equation
(r_{3})² + 2r_{1}r_{3} - x² = 0. |
The equation has two real roots, a negative one and a positive one, the latter being
r_{3} = - r_{1} + √(r_{1}² + x²) = 0. |
Substituting r_{1} from (1) gives
(2) | r_{3} = x² / r. |
Next in line is r_{4}:
for which the Pythagorean theorem provides
(r - r_{1} - r_{4})² + x² = (r_{1} + r_{4})². |
One simplification is immediate:
r² + x² = 2r(r_{1} + r_{4}), |
Substituting r_{1} from (1) gives
r² + x² = r² - x² + 2r·r_{4}, |
from which
(2) | r_{4} = x² / r. |
Thus we see that indeed r_{3} = r_{4}.
When x = r/3, r_{3} = r_{4} = r/9. Note that, for this value of x, r_{1} = 4r/9. Also, circles O_{1}(r_{1}) and O'_{1}(r_{1}) have centers 2x = 2r/3 apart giving the width of the lens shaped intersection as
2r_{1} - 2x = 8r/9 - 2r/3 = 2r/9, |
implying r_{5} = r/9 = r_{3} = r_{4}, for x = r/3.
The sangaku ends here, but having a dynamic applet invites an investigation. If the three equal circles grow while the quadruplets become small, the former eventually form 5 regions, into which one may want to inscribed circles:
When the five circles are equal, their common radius is r/5; obviously. More can be said, viz., the outer four circles are always equal. Furthermore, their radius is exactly r_{1} increasing the number of equal circles to 8. The proof is not difficult and is left as an exercise.
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- H. Fukagawa, A. Rothman, Sacred Geometry: Japanese Temple Geometry, Princeton University Press, 2008, p. 101
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Copyright © 1996-2018 Alexander Bogomolny
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