A Sangaku Follow-Up on an Archimedes' Lemma

The following Sangaku problem sports a publication trail:

The problem has been included in Fukagawa and Pedoe's Japanese Temple Geometry Problems, offered by Crux Mathematicorum, 1987, 194, and appeared in R. Honsberger's collection From Erdös To Kiev with two solutions from the Crux. The first two solutions below, have been adapted from Honsberger's book.

Solution 1

This solution is by Dan Sokolowsky, Williamsburg, VA.

Start by drawing PXQ perpendicular to L at X. Since L is a common tangent to A and B, PX and QX are diameters of A and B. If M and N are their points of tangency with C, angles at M and N subtended by PX and QX are right:

XMP = XNS = 90o.

Extending MX and MP to meet C in R and S, we get a diameter RS (since RS subtends right angles in C.)

Now, the centers U and V of A and C are in line with the point of tangency M, and it is clear that triangles MUX and MVS are isosceles. Since they share a common base angle at M, the base angles at X and S are also equal, and the lines PXQ and RS are parallel. Thus RS is perpendicular to L.

Similarly, extending NX and NQ we'll get a diameter perpendicular to L. Therefore, R lies on NX and S on NQ.

The next step is to extend MR and NS beyond M and N to meet in T. In ΔRST, RN and SM are altitudes, so that X is the orthocenter. TX is then the third altitude and is perpendicular to RS. But L is a line through X perpendicular to RS, which leads us to conclude that T lies on L.

The final step is to consider two pairs of similar triangles (TXP, TWR) and (TXQ, TWS):

so that

Since the right side here is independent of X, so is the left side. The conclusion follows.

I have only one remark concerning this proof. We may have started with constructing RS, a diameter perpendicular to PXQ. Then PR passes through M (and QS through N) by Lemma 1 from Archimedes' Book of Lemmas. (The situation has been explored elsewhere.)

Solution 2

This solution is by Sam Baethge, San Antonio, TX. It's a straightforward application of a theorem attributed to another famous Greek mathematician.

Let p and d be the horizontal and vertical components of VX. And let a, b and R be the radii of A, B, and C, respectively. By the Pythagorean theorem,

These are solved for a and b to get

from which a/b = (R - d)/(R + d) independent of X.

This unexciting solution proved extremely useful in constructing the diagram and, ultimately, writing the applet illustration.

Solution 3

This simple solution that is due to Michel Cabart employs inversion which I am not sure has been available to the Japanese during the period of seclusion.

Trace the tangents D₁ and D₂ to circle C in R and S. Now transform the figure by the inversion with center X, and power equal to the power of X relative to C. The latter equals k = XM·XS = XN·XR. In particular this means that M and X are inverse images of each other, and so are N and R.

Circle A is transformed into a straight line parallel to L through the image of M, which is D₂. Similarly, B is mapped onto D₁.

The inverse of P stays on the perpendicular to L through X, i.e., PQ, but also lies on the image of A, i.e. D₂. Hence the distance from X to the image of P is R + d. The distance from P itself to X is of course 2a. Thus 2a·(R + d) = k. Similarly, for the inverse of Q, 2b·(R - d) = k. Taking the ratio, we have the desired a/b = (R - d)/(R + d).

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7. A Geometric Mean Sangaku
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10. A Sangaku: Two Unrelated Circles
11. A Sangaku by a Teen
12. A Sangaku Follow-Up on an Archimedes' Lemma
13. A Sangaku with an Egyptian Attachment
14. A Sangaku with Many Circles and Some
15. A Sushi Morsel
16. An Old Japanese Theorem
17. Archimedes Twins in the Edo Period
18. Arithmetic Mean Sangaku
19. Bottema Shatters Japan's Seclusion
20. Circles and Semicircles in Rectangle
21. Circles in a Circular Segment
22. Circles Lined on the Legs of a Right Triangle
23. Equal Incircles Theorem
24. Equilateral Triangle, Straight Line and Tangent Circles
25. Equilateral Triangles and Incircles in a Square
26. Five Incircles in a Square
27. Four Hinged Squares
28. Four Incircles in Equilateral Triangle
29. Gion Shrine Problem
30. Harmonic Mean Sangaku
31. Heron's Problem
32. In the Wasan Spirit
33. Incenters in Cyclic Quadrilateral
34. Japanese Art and Mathematics
35. Malfatti's Problem
36. Maximal Properties of the Pythagorean Relation
37. Neuberg Sangaku
38. Out of Pentagon Sangaku
39. Peacock Tail Sangaku
40. Pentagon Proportions Sangaku
41. Pythagoras and Vecten Break Japan's Isolation
42. Radius of a Circle by Paper Folding
43. Review of Sacred Mathematics
44. Sangaku à la V. Thebault
45. Sangaku and The Egyptian Triangle
46. Sangaku in a Square
47. Sangaku Iterations, Is it Wasan?
48. Sangaku with 8 Circles
49. Sangaku with Three Mixtilinear Circles
50. Sangaku with Versines
51. Sangakus with a Mixtilinear Circle
52. Sequences of Touching Circles
53. Square and Circle in a Gothic Cupola
54. Tangent Circles and an Isosceles Triangle
55. The Squinting Eyes Theorem
56. Steiner's Sangaku
57. Three Incircles In a Right Triangle
58. Three Squares and Two Ellipses
59. Three Tangent Circles Sangaku
60. Triangles, Squares and Areas from Temple Geometry
61. Two Arbelos, Two Chains
62. Two Circles in an Angle
63. Two Sangaku with Equal Incircles

References

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4
2. R. Honsberger, From Erdös To Kiev, MAA, 1996.

Copyright © 1996-2009 Alexander Bogomolny