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