Arithmetic Mean Sangaku: What Is This About?
A Mathematical Droodle
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Solution
Sangaku
- Sangaku: Reflections on the Phenomenon
- Critique of My View and a Response
- 1 + 27 = 12 + 16 Sangaku
- 3-4-5 Triangle by a Kid
- 7 = 2 + 5 Sangaku
- A 49th Degree Challenge
- A Geometric Mean Sangaku
- A Hard but Important Sangaku
- A Sangaku: Two Unrelated Circles
- A Sangaku by a Teen
- A Sangaku Follow-Up on an Archimedes' Lemma
- A Sangaku with an Egyptian Attachment
- A Sangaku with Many Circles and Some
- An Old Japanese Theorem
- Archimedes Twins in the Edo Period
- Arithmetic Mean Sangaku
- Bottema Shatters Japan's Seclusion
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
- Circles Lined on the Legs of a Right Triangle
- Equal Incircles Theorem
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
- Five Incircles in a Square
- Four Hinged Squares
- Four Incircles in Equilateral Triangle
- Gion Shrine Problem
- Harmonic Mean Sangaku
- Heron's Problem
- In the Wasan Spirit
- Incenters in Cyclic Quadrilateral
- Japanese Art and Mathematics
- Malfatti's Problem
- Maximal Properties of the Pythagorean Relation
- Neuberg Sangaku
- Out of Pentagon Sangaku
- Peacock Tail Sangaku
- Pentagon Proportions Sangaku
- Pythagoras and Vecten Break Japan's Isolation
- Radius of a Circle by Paper Folding
- Review of Sacred Mathematics
- Sangaku ŕ la V. Thebault
- Sangaku and The Egyptian Triangle
- Sangaku in a Square
- Sangaku Iterations, Is it Wasan?
- Sangaku with 8 Circles
- Sangaku with Three Mixtilinear Circles
- Sangaku with Versines
- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
- Square and Circle in a Gothic Cupola
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Steiner's Sangaku
- Three Incircles In a Right Triangle
- Three Squares and Two Ellipses
- Three Tangent Circles Sangaku
- Triangles, Squares and Areas from Temple Geometry
- Two Arbelos, Two Chains
- Two Circles in an Angle
Copyright © 1996-2008 Alexander Bogomolny
The applet is hopefully suggestive of the following sangaku [Bicycle, #39]:
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Points C and D lie on a circle with diameter AB, CD AB. One circle is inscribed into ΔABC. Two more circles inscribed on the other side of AB in the curvilinear triangles formed by AB, CD and the given circle. Prove that the radius of the former is the average of the radii of the latter. Moreover, the lines through the centers of the latter circles are tangent to the former circle.
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We apply the Pythagorean theorem twice to the right triangles with hypotenuses joining the centers of the circles, as shown:
Let the distance from center O of the circumcircle to CD be d, the other centers I, P, Q, and the radii of the circles be R, r, rL, rR. Then the Pythagorean theorem gives
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(r + d)2 + r2 = (R - r)2, for r = rR and
(r - d)2 + r2 = (R - r)2, for r = rL,
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From which we express rR and rL:
| (1) |
rR = -(R + d) + √2R(R + d) and
rL = -(R - d) + √2R(R - d).
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which shows that the arithmetic mean of the two equals
| (2) |
(rR + rL)/2 = -R + (√2R(R + d) + √2R(R - d) )/2.
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One the other hand, we need to find the inradius r of the right ΔABC. For any right triangle with sides a, b and hypotenuse c, the inradius r can be found from
In our case, c = 2R, while a and b can again be found with the help of the Pythagorean theorem. It's not hard to see that one of them is √2R(R + d) whereas the other is √2R(R - d), which conforms with (2) and shows that the horizontal distance between P and Q is 2r. We may rewrite (1) as
| (1') |
rR = -(R + d) + b
rL = -(R - d) + a.
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Thus ST = 2r, where S and T are the points of tangency of (P) and (Q) with AB. In particular,
| (3) |
r = (rR + rL)/2 = -R + (a + b)/2.
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What remains to show is that the midpoint of ST is exactly the point of tangency W of the incircle (I) with AB, i.e. SW = TW = r.
Let's compute
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| TW | = BW - BT |
| | = (a - r) - (R - d - rR) | | |
| | = (a - r) - (2R - b) | | (by (1')) |
| | = (a+b) - r - 2R | | |
| | = (a+b) - 2R - r | | |
| | = 2r - r | | (by (3)) |
| | = r. | | |
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And we are done.
(This sangaku has a nice generalization.)
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
- J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996
Copyright © 1996-2008 Alexander Bogomolny
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